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‎2018.11.18/大腩肚.md

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public List<List<Integer>> threeSum(int[] nums) {
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final int length = nums.length;
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List<List<Integer>> result = new ArrayList<>();
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HashMap<Integer, int[]> hashMap = new HashMap<Integer, int[]>();
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if (length < 3) return result;
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Arrays.sort(nums);
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for (int i = 0; i < length - 2; i++) {
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hashMap.clear();
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if (i == 0 || nums[i] > nums[i - 1]) {
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for (int j = i + 1; j < length; j++) {
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if (hashMap.containsKey(nums[j])) {
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ArrayList<Integer> elem = new ArrayList<Integer>(3);
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elem.add(hashMap.get(nums[j])[0]);
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elem.add(hashMap.get(nums[j])[1]);
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elem.add(nums[j]);
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result.add(elem);
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while (j < (length - 1) && nums[j] == nums[j + 1]) j++;
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} else {
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int[] temp = new int[2];
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temp[0] = nums[i];
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temp[1] = nums[j];
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hashMap.put(0 - (nums[i] + nums[j]), temp);
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}
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}
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}
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}
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return result;
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}

‎2018.11.27-leetcode125/暮成雪.md

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class Solution {
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public boolean isPalindrome(String s) {
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s = s.replaceAll("[^a-zA-Z]", "").toLowerCase();
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s = s.toLowerCase();
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char [] charArray = s.toCharArray();
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String temp = "";
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for(int i=0;i<charArray.length;i++)
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{
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if( ((int)charArray[i] >= 48 && (int)charArray[i] <= 57) || ((int)charArray[i] >= 97 && (int)charArray[i] <= 122))
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{
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temp += charArray[i];
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}
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}
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char [] resultArray = temp.toCharArray();
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int begin = 0;int end = resultArray.length - 1;
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while(begin < end)
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{
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if(resultArray[begin] == resultArray[end])
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{
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begin++;
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end--;
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}else{
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return false;
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}
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}
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return true;
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}
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}

‎2018.11.29-leetcode443/我爱吃瓜子.md

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#leetcode 443
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、、、
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class Solution {
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public int compress(char[] chars) {
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int count =1;
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int index= 0;
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for(int i=0;i<chars.length;i++){
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//判断是否最后一个 或者 判断是否重复的最后一个
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if(i+1 == chars.length || chars[i]!=chars[i+1]){
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//将字母赋值到结果字符数组中
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chars[index++]=chars[i];
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//判断是否需要将数量赋值到数组中
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if(count>1){
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//int转为String
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String temp =String.valueOf(count);
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//将数量赋值到结果数组中
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for(int z=0;z<temp.length();z++){
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//获取字符串中某个下标的字符
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chars[index++] =temp.charAt(z);
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}
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//初始化数量
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count=1;
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}
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}else{ //字符重复中
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count++;
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}
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}
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return index;
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}
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}
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、、、

‎2018.11.30-leetcode890/Avalon.md

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class Solution {
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public List<String> findAndReplacePattern(String[] words, String pattern) {
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List<String> wlist = new ArrayList<>();
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int len = words.length;
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String template = getTemplate(pattern);
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for (int i=0;i<len;i++){
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if (words[i].length()==pattern.length()){
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String temp = getTemplate(words[i]);
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if (temp.equals(template))
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wlist.add(words[i]);
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}
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}
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return wlist;
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}
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private static String getTemplate(String pattern){
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char[] cs = pattern.toCharArray();
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List<Character> cl = new ArrayList<>();
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String result="";
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int len = cs.length;
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for (int i=0;i<len;i++){
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if (cl.contains(cs[i]))
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result+=String.valueOf(cl.indexOf(cs[i])+1);
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else{
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cl.add(cs[i]);
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result += String.valueOf(cl.size());
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}
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}
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return result;
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}
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}

‎2018.11.30-leetcode890/Saul.md

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```java
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public class Solution {
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public static List<String> findAndReplacePattern(String[] words, String pattern) {
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List<String> result = new ArrayList<>();
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boolean flag = false;
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for (String word : words) {
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if (word.length() != pattern.length()) {
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continue;
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}
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flag = false;
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char[] wordChars = word.toCharArray();
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char[] patternChars = pattern.toCharArray();
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//比较wordChars第i个位置与第j个位置的值，如果相同，那么patternChars第i个位置与第j个位置的值也要相同
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//如果wordChars第i个位置与第j个位置的值不同，那么patternChars第i个位置与第j个位置的值也要不同
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for (int i = 0; i < wordChars.length ; i++) {
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for (int j = i + 1; j < wordChars.length; j++) {
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if (wordChars[i] == wordChars[j]) {
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if (patternChars[i] != patternChars[j]) {
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flag = true;
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break;
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}
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}
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if (wordChars[i] != wordChars[j]) {
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if (patternChars[i] == patternChars[j]) {
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flag = true;
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break;
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}
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}
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}
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}
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if (!flag) {
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result.add(word);
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}
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}
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return result;
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}
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}
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```

‎2018.11.30-leetcode890/kiritocly.md

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private static List<String> solution(String[] words, String pattern) {
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//定义一个list存放结果字符串
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List<String> result = new ArrayList<>();
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char[] patternArray = pattern.toCharArray();
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//遍历字符串
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for (int i = 0; i < words.length; i++) {
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//匹配字符串
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char[] wordArray = words[i].toCharArray();
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//是否匹配
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boolean flag = true;
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//长度相同才能匹配
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if (wordArray.length == patternArray.length) {
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//定义两个hashMap存放映射:words中的单词与pattern
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Map<Character, Character> hashMap1 = new HashMap<>();
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Map<Character, Character> hashMap2 = new HashMap<>();
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//逐个字符匹配
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for (int j = 0; j < patternArray.length; j++) {
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if (!hashMap1.containsKey(patternArray[j])) {
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//如果不存在字符则向hashMap1存放映射关系对
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hashMap1.put(patternArray[j], wordArray[j]);
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//判断hashMap2中是否存在反向映射关系对
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if (!hashMap2.containsKey(wordArray[j])) {
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hashMap2.put(wordArray[j], patternArray[j]);
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} else {
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if (hashMap2.get(wordArray[j]) != patternArray[j]) {
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//存在反向字符且当前判断的字符不相同
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flag = false;
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break;
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}
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}
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} else {
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if (hashMap1.get(patternArray[j]) != wordArray[j]) {
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flag = false;
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break;
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}
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}
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}
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}
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if (flag) {
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result.add(words[i]);
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}
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}
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return result;
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}

‎2018.11.30-leetcode890/家

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```
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func findAndReplacePattern(words []string, pattern string) []string {
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l := len(pattern)
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a := map[string]int{}
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arr := []int{}
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for i:=0;i<=l-1;i++{
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if _,ok := a[string(pattern[i])];!ok {
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a[string(pattern[i])] = len(a)
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}
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arr = append(arr,a[string(pattern[i])])
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}
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newstr := []string{}
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for _,v := range words {
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a = map[string]int{}
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flag := true
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for i:=0;i<len(v);i++{
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if _,ok := a[string(v[i])];!ok {
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a[string(v[i])] = len(a)
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}
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if a[string(v[i])] != arr[i]{
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flag = false
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break;
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}
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}
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if flag {newstr = append(newstr,v)}
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}
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return newstr
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}
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```

‎2018.11.30-leetcode890/我爱吃瓜子

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#leetcode 890
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```
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//思路是保证对应下标的权重是一样的，只需要比较权重就可以判断是否符合规则
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class Solution {
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public static List<String> findAndReplacePattern(String[] words, String pattern) {
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List<String> result = new ArrayList<>();//返回结果
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char [] patternArray = pattern.toCharArray();//字符串转字符数组
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for(int i=0;i<words.length;i++)//遍历每一个words中的字符串
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{
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char [] wordArray = words[i].toCharArray();
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boolean flag = true;//标记，为true说明匹配
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if(wordArray.length == patternArray.length)//匹配肯定长度要相等
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{
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HashMap<Character,Character> judge = new HashMap<>();//hashmap1
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HashMap<Character,Character> judge2 = new HashMap<>();//反向比较hashmap2
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for(int j=0;j<pattern.length();j++)//遍历模式串中的每一个字符
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{
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if(judge.containsKey(patternArray[j]) == false)
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{//判断key中是否存在这个字符，不存在hashmap1就添加key-value
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judge.put(patternArray[j],wordArray[j]);
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if(judge2.containsKey(wordArray[j]) == false)
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{//这个if else的作用就是思路中的hashmap2的作用了
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judge2.put(wordArray[j],patternArray[j]);//不存在就添加
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}else {
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if(judge2.get(wordArray[j]) != patternArray[j]){
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flag = false;//存在的话就去判断是不是相等的，不相等就不匹配
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break;
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}
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}
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}else {//存在的话 去hashmap中找这个字符与现有的字符是否相等
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//不相等，说明存在了（a-c a-b 一个a对应两个字符）直接返回false,不匹配
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if(judge.get(patternArray[j]) != wordArray[j])
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{
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flag = false;
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break;
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}
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}
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}
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if(flag)
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{
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result.add(words[i]);
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}
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}
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}
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return result;
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}
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}
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```

‎2018.12.04-leetcode101/101. 对称二叉树

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给定一个二叉树，检查它是否是镜像对称的。
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例如，二叉树 [1,2,2,3,4,4,3] 是对称的。
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1
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/ \
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2 2
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/ \ / \
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3 4 4 3
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但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:
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1
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/ \
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2 2
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\ \
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3 3
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说明:
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如果你可以运用递归和迭代两种方法解决这个问题，会很加分。

‎2018.12.05-leetcode102/二叉树的层次遍历.md

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定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
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例如:
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给定二叉树: [3,9,20,null,null,15,7],
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3
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/ \
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9 20
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/ \
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15 7
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返回其层次遍历结果：
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[
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[3],
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[9,20],
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[15,7]
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]

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