Leetcode 374. Guess Number Higher or Lower

374. Guess Number Higher or Lower

• Total Accepted: 7396
• Total Submissions: 23391
• Difficulty: Easy

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：二分查找。这里的二分查找模板可以推广。下面有2个模板，注意各自的应用场景。

代码：

这个是在左闭右闭区间[low,high]中查找元素值>=查找值的第1个元素的位置，这里也考虑high位置和查找值的大小。

// 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,high=n,mid;
        while(low<=high){
            mid=low+(high-low)/2;
            if(guess(mid)==1){
                low=mid+1;
            }
            else{
                high=mid-1;
            }
        }
        return low;
    }
};

这个是在左闭右开区间[low,high)中查找元素值>=查找值的第1个元素的位置，这里不考虑high位置和查找值的大小。相当于STL的函数lower_bound()。

本题有些特殊，就算high所在位置不被比较，如果排查了high之前所有位置和查找值的大小，最后low和high都停留在同1个位置，这时high虽然不被比较，但肯定就是high位置元素==查找值。例如在[1,2,3]中查找3。

// 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,high=n,mid;
        while(low<high){
            mid=low+(high-low)/2;
            if(guess(mid)==1){
                low=mid+1;
            }
            else{
                high=mid;
            }
        }
        return low;
    }
};