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.
int guess(int num);//api函数
int guessNumber(int n)
1、给定一个整数 n，从1到n中挑一个数出来，让你猜是哪个数。每次你猜一个数，调用一次api函数，返回0就代表刚好是这个数，你猜对了！如果返回-1，就表示你猜的数太大了，实际的数比这个小。如果返回1，就表示你猜的数太小了，实际的数比这个大。要求最终返回对的那个数。
2、这道题很明显是一道二分查找的题目，手工写一个二分查找的代码。
代码如下：（附详解）
int guess(int num);
int guessNumber(int n)
{
int low=1,high=n,mid,t;
while(low<=high)
{
mid=low+(high-low)/2;//写成 mid=(low+high)/2 可能会上溢
t=guess(mid);
if(t==0)
return mid;
else if(t==-1)//实际的数比mid小
high=mid-1;
else //实际的数比mid大
low=mid+1;
}
}
上述代码实测2ms，beats 100.00% of cpp submissions。
leetcode-374-Guess Number Higher or Lower（二分查找）
原文：https://www.cnblogs.com/king-3/p/9125248.html