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，虽然是道 Easy 题，无脑线性遍历还是会超时 Time Limit Exceeded，所以更快速的方法就是折半搜索法，原理很简单，属于博主之前的总结帖 LeetCode Binary Search Summary 二分搜索法小结 中的第四类-用子函数当作判断关系，参见代码如下：
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;
if (t == 1) left = mid + 1;
else right = mid;
}
return left;
}
};
Github 同步地址：
https://github.com/grandyang/leetcode/issues/374
类似题目：
Guess Number Higher or Lower II
参考资料：
https://leetcode.com/problems/guess-number-higher-or-lower/
https://leetcode.com/problems/guess-number-higher-or-lower/discuss/84664/0ms-c%2B%2B-binary-search
LeetCode All in One 题目讲解汇总(持续更新中…)
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