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.
这道题是一道典型的猜价格的问题，根据对方说高了还是低了来缩小范围，最简单快速的方法就是折半搜索法，原理很简单，参见代码如下：
// 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) { if (guess(n) == 0) return n; int left = 1, right = n; while (left < right) { int mid = left + (right - left) / 2, t = guess(mid); if (t == 0) return mid; else if (t == 1) left = mid; else right = mid; } return left; } };
本文转自博客园Grandyang的博客，原文链接：猜数字大小[LeetCode] Guess Number Higher or Lower ，如需转载请自行联系原博主。
