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Merge pull request gzc426#7 from gzc426/master
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‎.idea/vcs.xml

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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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```

‎2018.11.25-leetcode80/tonythecyclist.md renamed to ‎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){

‎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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```

‎2018.11.26-leetcode11/Ostrichcrab.md

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```
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class Solution {
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public:
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int max(int a, int b){return a>b?a:b;}
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int min(int a, int b){return a<b?a:b;}
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int maxArea(vector<int>& height) {
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int ans = 0;
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int l = 0, r = height.size() - 1;
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ans = max(ans,min(height[l],height[r])*(r-l));
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while(l<r){
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if(height[l]>height[r]){
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r--;
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ans = max(ans,min(height[l],height[r])*(r-l));
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}else{
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l++;
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ans = max(ans,min(height[l],height[r])*(r-l));
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}
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}
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return ans;
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}
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};
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```

‎2018.11.26-leetcode11/tonythecyclist.md renamed to ‎2018.11.26-leetcode11/Tony the Cyclist.md

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‎2018.11.25-leetcode11/str818.md renamed to ‎2018.11.26-leetcode11/str818.md

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‎2018.11.26-leetcode11/syuan.md

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##### 题目
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```
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给定 n 个非负整数 a1，a2，...，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0)。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。
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说明：你不能倾斜容器，且 n 的值至少为 2。
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```
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![微信截图_20181127122321](2DF0C29B574D4A30BFAB06ABDF185FA9)
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```
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图中垂直线代表输入数组 [1,8,6,2,5,4,8,3,7]。在此情况下，容器能够容纳水（表示为蓝色部分）的最大值为 49。
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示例:
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输入: [1,8,6,2,5,4,8,3,7]
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输出: 49
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```
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##### 思路
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双指针
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##### 代码
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```
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class Solution {
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public int maxArea(int[] height){
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int maxarea=0;
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int i=0;
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int j=height.length-1;
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while(i<j){
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maxarea=Math.max(maxarea,Math.min(height[i],height[j])*(j-i));
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if(height[i]<height[j]){
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i++;
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}else{
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j--;
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}
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}
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return maxarea;
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}
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}
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```

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