LeetCode 374. Guess Number Higher or Lower

题目：
We are playing the Guess Game. The game is as follows:
I pick a number from 1 to n. You have to guess which number I picked.
Every time you guess wrong, I’ll tell you whether the number is higher or lower.

You call a pre-defined API guess(int num) which returns 3 possible results (-1, 1, or 0):
-1 : My number is lower
1 : My number is higher
0 : Congrats! You got it!

Example:
n = 10, I pick 6.

Return 6.
思路：
如果给的值比要找的大，guess返回-1；如果给的值比要找的小，guess返回1；如果就是要找的值，返回0.
利用二分查找来找这个值，先将1赋给low,n赋给high，num其实是mid，不要写(high+mid)/2，先进行一次判断num是否为要找的值，返回值赋给res。当res不为0，那么其只能为1或者-1，如果为1，即给的数比较小，要改变low为num+1，更新num,再做判断此时的num是否是要找的值；如果为-1时即给的数比较大，要改变high为num-1，更新num,再做判断此时的num是否是要找的值。循环结束后，num即为要找的值。

代码：

// Forward declaration of guess API.
// @param num, your guess
// @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
int guess(int num);class Solution {
public:int guessNumber(int n) {int low=1;int high=n;//先将1赋给low,n赋给highint num=(high-low)/2+low;//num其实是mid，不要写(high+mid)/2int res=guess(num);//先进行一次判断num是否为要找的值，返回值赋给reswhile(res){//如果res不为0，那么其只能为1或者-1if(res==1){//如果为1，即给的数比较小，要改变low为num+1low=num+1;num=(high-low)/2+low;//更新num,再做判断此时的num是否是要找的值res=guess(num);}else{//如果为-1时即给的数比较大，要改变high为num-1high=num-1;num=(high-low)/2+low;//更新num,再做判断此时的num是否是要找的值res=guess(num);}}return num;//循环结束后，num即为要找的值}
};

输出结果： 0ms

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