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Commit 1924692

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Update README.md
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‎README.md

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@@ -33,7 +33,7 @@ LeetCode 是一个非常棒的 OJ(Online Judge)平台,收集了许多公
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> Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring. For "(()", the longest valid parentheses substring is "()", which has length = 2. Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4.
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```
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```C#
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//栈的典型应用
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//能想到栈,但是能想到Push字符索引,真的不是很容易(这道题一般的想法是存储‘(’ ‘)’这些符号)
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public int LongestValidParentheses(string s)
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return longest;
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}
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```
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|ID|Tags|Solution|
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|--|----|--------|
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|3|String;|[Longest Substring Without Repeating Characters](http://blog.csdn.net/daigualu/article/details/73105674)|
@@ -194,7 +195,7 @@ return heapq.nlargest(k, num_count, key=lambda x: num_count[x])
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而用 C++ 实现的话,代码会多很多,带来的好处就是速度的飞跃。具体代码在 这里 ,建立大小为 k 的小顶堆,每次进堆时和堆顶进行比较,核心代码如下:
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```
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```c++
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// Build the min-heap with size k.
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for(auto it = num_count.begin(); it != num_count.end(); it++){
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if(frequent_heap.size() < k){
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如果用 C++ 的话,相信很多人也能避开求中间值的整型溢出的坑: int mid = low + (high - low) / 2; ,于是写出下面的代码:
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```
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```c++
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int low = 0, high = x;
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while(low <= high){
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// int mid = (low+high) / 2, may overflow.
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除了臭名昭著的整型溢出问题,c++ 和 python 在位运算上也有着一点不同。以 371 Sum of Two Integers 为例,不用 +, - 实现 int 型的加法 int getSum(int a, int b) 。其实就是模拟计算机内部加法的实现,很明显是一个位运算的问题,c++实现起来比较简单,如下:
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```
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```c++
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int getSum(int a, int b) {
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if(b==0){
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return a;
@@ -270,7 +271,7 @@ StefanPochmann 也不厌其烦地给出了自己的答案:
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<center>
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![这里写图片描述](http://img.blog.csdn.net/20170610204936185?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvZGFpZ3VhbHU=/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast)
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<h4>已完成题目列表</h4>
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## 已完成题目列表
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|ID|Tags|Solution|
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|--|----|--------|
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|1|Array;HashTable;|[Two Sum](http://blog.csdn.net/daigualu/article/details/68957096)|

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