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‎.idea/vcs.xml

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‎2018.11.19-leetcode15/Istoryforever.md

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leetcode of 15
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```
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class Solution {
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public:
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vector<vector<int>> threeSum(vector<int>& nums) {
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vector<vector<int>> res;
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if(nums.size()==0) return res;
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sort(nums.begin(),nums.end());
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int i=0;
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int p=0;
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int q=0;
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while(i<nums.size()-2){
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if(i>0 &&nums[i-1]==nums[i]){
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i++;
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continue;
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}
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if(nums[i]>0) break;
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int diff=0-nums[i];
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p=i+1;
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q=nums.size()-1;
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while(p<q){
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if(nums[p]+nums[q]<diff && p<q) p++;
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if(nums[p]+nums[q]>diff && p<q) q--;
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else if(p<q && nums[p]+nums[q]==diff){
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vector<int>temp;
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temp.push_back(nums[i]);
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temp.push_back(nums[p]);
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temp.push_back(nums[q]);
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res.push_back(temp);
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p++;
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q--;
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while(p<q &&nums[p]==nums[p-1]){
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p++;
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}
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while(p<q &&nums[q]==nums[q+1]){
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q--;
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}
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}
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}
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i++;
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}
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return res;
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}
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};
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```

‎2018.11.24-leetcode28/Tony the Cyclist.md

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```
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class Solution {
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public int strStr(String haystack, String needle) {
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int len = haystack.length();
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int sublen = needle.length();
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for (int i = 0; i < len - sublen + 1; i++){
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if (haystack.substring(i, i + sublen).equals(needle)){
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return i;
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}
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}
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return -1;
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}
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}
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```

‎2018.11.24-leetcode28/sourcema.md

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# LeetCode 28
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public static int strStr(String haystack, String needle) {
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if (needle.length() == 0) {
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return 0;
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}
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if (haystack.length() == 0 || haystack == null || needle.length() > haystack.length()) {
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return -1;
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}
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int hStart,nStart;
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for (int i = 0; i < haystack.length(); i++) {
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hStart=i;
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nStart=0;
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while (nStart < needle.length() && hStart < haystack.length()) {
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if (haystack.charAt(hStart) == needle.charAt(nStart)) {
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hStart++;
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nStart++;
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} else {
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break;
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}
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if (nStart == needle.length()) {
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return i;
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}
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}
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}
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return -1;
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}

‎2018.11.24-leetcode28/syuan.md

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```
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class Solution {
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public int strStr(String haystack, String needle) {
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if(needle.length() == 0)
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return 0;
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char [] charArrayhaystack = haystack.toCharArray();
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char [] charArrayneedle = needle.toCharArray();
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for(int i =0;i<charArrayhaystack.length;i++)
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{
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if(charArrayhaystack[i] == charArrayneedle[0])
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{
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int j = i + 1;
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int k = 1;
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for(k=1;k<charArrayneedle.length && j < charArrayhaystack.length;k++)
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{
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if(charArrayhaystack[j] == charArrayneedle[k])
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{
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j++;
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}else{
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break;
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}
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}
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if(k == charArrayneedle.length)
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return i;
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}
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}
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return -1;
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}
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}
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```

‎2018.11.24-leetcode28/大魔王爱穿孖烟筒.md

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class Solution {
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public int strStr(String haystack, String needle) {
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if(needle.length()==0){
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return 0;
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}
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char [] hay=haystack.toCharArray();
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char [] nde=needle.toCharArray();
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for(int i=0;i<hay.length;i++){
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if(hay[i]==nde[0]){ //遇到相同的情况
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int j=i+1;//指针j比较i的下一个元素与needle是否相同
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int k=1; //指针k遍历needle数组，从第二个开始比较，第一个已经判断相同的情况下才进入的。
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for(;k<nde.length&&j<hay.length;k++){
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if(hay[j]==nde[k]){//如果下一个元素还相同
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j++; //则比较haystack的下一个
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}else{
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break;//很快遇到不同的则退出。
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}
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}
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if(k==nde.length){
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return i;
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}
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}
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}
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return -1;
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}
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}

‎2018.11.25-leetcode80/Istoryforever.md

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leetcode number 80
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```
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class Solution {
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public:
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void swap(int &a,int &b){
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int c = a;
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a = b;
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b = a;
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}
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int removeDuplicates(vector<int>& nums) {
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int j = -1;
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int count = 0;
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for(int i = 0;i < nums.size();i++){
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if(i != 0 && nums[i] == nums[i-1]) {
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if(count >= 1) continue;
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++count;
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}
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else
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count = 0;
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nums[++j]=nums[i];
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}
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return j+1;
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}
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};
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```

‎2018.11.25-leetcode80/Tony the Cyclist.md

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```
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class Solution {
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public int removeDuplicates(int[] nums) {
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if (nums.length < 3){
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return nums.length;
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}
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int pos = 2;
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for(int i = 2; i < nums.length; i++){
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if(nums[i]!=nums[pos-2])
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nums[pos++]=nums[i];
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}
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return pos;
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}
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}
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```

‎2018.11.25-leetcode80/syuan.md

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##### 题目
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```
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给定一个排序数组，你需要在原地删除重复出现的元素，使得每个元素最多出现两次，返回移除后数组的新长度。
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不要使用额外的数组空间，你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。
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示例 1:
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给定 nums = [1,1,1,2,2,3],
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函数应返回新长度 length = 5, 并且原数组的前五个元素被修改为 1, 1, 2, 2, 3 。
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你不需要考虑数组中超出新长度后面的元素。
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示例 2:
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给定 nums = [0,0,1,1,1,1,2,3,3],
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函数应返回新长度 length = 7, 并且原数组的前五个元素被修改为 0, 0, 1, 1, 2, 3, 3 。
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你不需要考虑数组中超出新长度后面的元素。
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说明:
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为什么返回数值是整数，但输出的答案是数组呢?
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请注意，输入数组是以“引用”方式传递的，这意味着在函数里修改输入数组对于调用者是可见的。
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你可以想象内部操作如下:
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// nums 是以“引用”方式传递的。也就是说，不对实参做任何拷贝
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int len = removeDuplicates(nums);
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// 在函数里修改输入数组对于调用者是可见的。
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// 根据你的函数返回的长度, 它会打印出数组中该长度范围内的所有元素。
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for (int i = 0; i < len; i++) {
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print(nums[i]);
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}
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```
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##### 代码
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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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if (nums.length==1) {
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return 1;
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}
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if (nums.length==0) {
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return 0;
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}
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if (nums.length==2) {
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return 2;
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}
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int i=0;
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int j=i;
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int tempCount=0;
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while(i<nums.length)
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{
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if (j<nums.length && nums[j]==nums[i]) {
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j++;
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tempCount++;
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}else{
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if (tempCount>=2) {
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count+=2;
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nums[i+1]=nums[i];
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i=i+2;
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}else{
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count++;
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i++;
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}
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if (j>=nums.length) {
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break;
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}
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nums[i]=nums[j];
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tempCount=0;
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}
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}
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return count;
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}
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}
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```

‎2018.11.25-leetcode80/大魔王爱穿孖烟筒.md

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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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if(nums.length==0){
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return 0;
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}
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if(nums.length==0){
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return 1;
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}
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int i=0;
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int j=i;//用来遍历数组
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int tempCount=0;
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while(i<nums.length){
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if(j<nums.length&&nums[j]==nums[i]){//固定i，j遍历，满足j位置的值和i相同，则j++；
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j++;
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tempCount++;
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}else{
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if(tempCount>=2){//统计每个位置数字出现的次数。
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count+=2;
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nums[i+1]=nums[i];//保存i的值，前后两个是相同的值，eg 1,1，2，。。
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i=i+2;//i来到下一个的下一个位置。
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}else{//相同的只有两个
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count++;
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i++;
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}
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if(j>=nums.length){
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break;
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}
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nums[i]=nums[j];//
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tempCount=0;//重新统计
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}
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}
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return count;
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}
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}

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