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Merge pull request gzc426#11 from gzc426/master
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‎2018.11.25-leetcode80/大魔王爱穿孖烟筒.md

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```
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class Solution {
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public int removeDuplicates(int[] nums) {
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int count=0;
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return count;
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}
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}
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```

‎2018.11.26-leetcode11/大魔王爱穿孖烟筒.md

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```
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class Solution {
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public int maxArea(int[] height) {
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int left=0;//从左边遍历
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int right=height.length-1;//从右边遍历
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int tempCha=0;
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int sum=0;
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int tempsum=0;
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while(left<right){//遍历到中间，两指针重合为止，求出沿途的面积。
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tempCha=right-left;//每次遍历记录临时长度差，也就是底边的长度
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if(height[right]>height[left]){
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tempsum=height[left]*tempCha; //短（短板）的边作为高，求出此时的面积
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left++;
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}else{
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tempsum=height[right]*tempCha;
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right--;
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}
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if(tempsum>sum){
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sum=tempsum;
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}
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}
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return sum;
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}
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}
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```

‎2018.11.27-leetcode125/大魔王爱穿孖烟筒.md

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```
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class Solution {
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public boolean isPalindrome(String s) {
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s = s.toLowerCase();
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char [] charArray = s.toCharArray();
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int begin = 0;
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int end = charArray.length - 1;
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while(begin < end)
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{
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if (!isChar(charArray[begin])){//不是数字和小字母
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begin++;
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continue;
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}
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if (!isChar(charArray[end])){//不是数字和小字母
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end--;
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continue;
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}
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if(charArray[begin] == charArray[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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public boolean isChar(char c){//判断是否为数字或者字母
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if (c<48||(c>57&&c<65)||(c>90&&c<97)||c>122){
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return false;
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}
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return true;
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}
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}
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```

‎2018.11.28-leetcode151/大魔王爱穿孖烟筒.md

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```
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public class Solution {
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public String reverseWords(String s) {
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if (s == null || s.length() == 0) {
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return "";
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}
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String[] array = s.split(" ");//按照空格划分字符串
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String str = "";
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for (int i = array.length - 1; i >= 0; i--) {
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if (!array[i].equals("")) {
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if (str.length() > 0) {
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str += " ";
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}
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str += array[i];//如果不是空字符串，就把截断的旧字符串添加到新字符串中。
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}
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}
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return str;
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}
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}
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```

‎2018.11.29-leetcode443/大魔王爱穿孖烟筒.md

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```
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class Solution {
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public int compress(char[] chars) {//思路有一点，写的时候就缺胳膊少腿，又来模仿群主代码
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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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if (i + 1 == chars.length || chars[i] != chars[i+1]) {
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chars[index++] = chars[i];//前后不同直接添加
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if (count > 1) {
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String temp = String.valueOf(count);//转换成字符串,也可以变成字符数组，然后遍历。
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for(int k=0;k<temp.length();k++)
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chars[index++] = temp.charAt(k);
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}
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count = 1;//重置
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}
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else {
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count++;//统计重复字符的个数
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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/大魔王爱穿孖烟筒.md

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```
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class Solution {
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public List<String> findAndReplacePattern(String[] words, String pattern) {
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List<String> list=new ArrayList<>();
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char [] pat=pattern.toCharArray();
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for(int i=0;i<words.length;i++){
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char [] sArray=words[i].toCharArray();
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boolean flag=true;
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if(sArray.length==pat.length) {
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HashMap<Character,Character> hash=new HashMap<>();
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HashMap<Character,Character> hash2=new HashMap<>();
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for(int j=0;j<pat.length;j++){
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if(hash.containsKey(pat[j])==false){//一个模式字符不存在的话，添加key和value
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hash.put(pat[j],sArray[j]);
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if(hash2.containsKey(sArray[j])==false){//考虑存在key和value重复的情况，使用另外一个map反向添加key和value。
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hash2.put(sArray[j],pat[j]);
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}else{//如果已经存在了
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if(hash2.get(sArray[j])!= pat[j]){
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flag=false;
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break;
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}
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}
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}else{//已经存在相应的key
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if(hash.get(pat[j])!= sArray[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){//如果模式对应，添加到list集合中
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list.add(words[i]);
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}
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}
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return list;
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}
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}
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```

‎2018.11.30-leetcode890/神秘的火柴人.md

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class Solution {
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public:
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vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
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vector<string> ret;
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int size = words.size();
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int n = pattern.length();
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for(int i=0; i<size; i++)
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{
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map<char, char> m;
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map<char, char> m1;
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int j = 0;
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for(j=0; j<n; j++)
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{
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if(m.find(pattern[j])==m.end())
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{
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m[pattern[j]] = words[i][j];
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}
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if(m1.find(words[i][j])==m1.end())
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{
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m1[words[i][j]] = pattern[j];
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}
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if(m.find(pattern[j])!=m.end() && m[pattern[j]] != words[i][j])
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break;
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if(m1.find(words[i][j])!=m.end() && m1[words[i][j]]!=pattern[j])
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break;
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}
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if(j == n)
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ret.push_back(words[i]);
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}
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return ret;
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}
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};

‎2018.12.04-leetcode101/Avalon.md

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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public boolean isSymmetric(TreeNode root) {
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if (root == null ){
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return true;
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}
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return isSymmetric(root.left, root.right);
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}
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public static boolean isSymmetric(TreeNode nodeA, TreeNode node_A){
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if (node_A == null && nodeA == null) {
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return true;
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}
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if (node_A != null && nodeA != null) {
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return node_A.val == nodeA.val && isSymmetric(nodeA.left, node_A.right) && isSymmetric(nodeA.right, node_A.left);
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}
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return false;
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}
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}

‎2018.12.04-leetcode101/MQQM.cpp

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/*
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题目：
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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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做法：
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给定对应位置上的两个节点，递归判断这两个节点是否一致。
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参考：
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https://www.cnblogs.com/xiaozhuyang/p/7376749.html
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*/
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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bool isSymmetric(TreeNode* root) {
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if( root == NULL ){
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return true;
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}
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return isEqual(root->left, root->right);
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}
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bool isEqual(TreeNode* left, TreeNode* right){//给定的两个节点已经是对应位置上的节点
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if( left == NULL && right == NULL ){//两个节点都是空
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return true;
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}
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if( left == NULL || right == NULL ){//有一个节点是空
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return false;
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}
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if( left->val != right->val ){//两个节点都存在，但是不相同
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return false;
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}
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return isEqual(left->left, right->right) && isEqual(left->right, right->left);
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}
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};

‎2018.12.04-leetcode101/marguerite.md

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101_(对称二叉树)Symmetric Tree
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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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输入：
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TreeNode* root：二叉树的根节点指针
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输出：
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bool：是否是对称二叉树
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输入: 给定二叉树: [1,2,2,3,4,4,3] ,
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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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输出: true
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solution1
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判断是否是镜像对称，重要的就是判断对应位置的2个结点的值是否相等。用双端队列来实现
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public boolean isSymmetric(TreeNode root){
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if(root == null){
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return true;
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}
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//队列做遍历
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Deque<TreeNode> deque = new LinkedList<TreeNode>();
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//根节点不用判断，直接加入左右孩子结点
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deque.addFirst(root.left);
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deque.addLast(root.right);
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TreeNode preNode = null;
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TreeNode postNode = null;
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while(!deque.isEmpty()){
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//取出列队两个元素
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preNode = deque.pollFirst();
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postNode = deque.pollLast();
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//如果当前结点是空结点，结束本次循环，即不需要检查后续结点
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if(preNode == null && postNode == null){
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continue;
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}
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//如果不对称返回false，
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if(preNode == null || postNode == null){
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return false;
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}
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if(preNode.val != postNode.val){
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return false;
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} else{
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deque.addFirst(preNode.right);//preNode右孩子入队
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deque.addFirst(preNode.left);//preNode左孩子入队
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deque.addLast(postNode.left);//postNode左孩子入队
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deque.addLast(postNode.right);//postNode右孩子入队
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}
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}
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return true;//都是对称的返回真
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}
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solution2
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public boolean isSysmmetric2(TreeNode root){
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if(root == null){
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return true;
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}
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return checkNodes(root.left,root.right);
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}
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public boolean checkNodes(TreeNode node1,TreeNode node2){
69+
if(node1 == null && node2 == null){
70+
return true;
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}
72+
if(node1 == null || node2 == null){
73+
return false;
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}
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if(node1.val != node2.val){
76+
return false;
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} else{
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return checkNodes(node1.left,node2.right) && checkNodes(node1.right,node2.left);
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}
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}
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