LEETCODE-- Guess Number Higher or Lower

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.
算法思想就是折半查找；
关于time limit exceeded:
当right的值位于上边界时
使用mid = (right + left) / 2;可能在计算right + left 时发生数值越界，导致超时；
使用mid = (right - left) / 2 + left;可以避免。

// Forward declaration of guess API.// @param num, your guess// @return -1 if my number is lower, 1 if my number is higher, otherwise return 0int guess(int num);class Solution {public:    int guessNumber(int n) {        int left = 1;        int right = n;        int mid = 0;        while(left <= right){            mid = (right - left) / 2 + left;            if(guess(mid) == 1)                left = mid + 1;            else if(guess(mid) == -1)                right = mid - 1;            else if(guess(mid) == 0)                return mid;        }        return left;    }