LEETCODE-- Guess Number Higher or Lower

    xiaoxiao2021-03-25  62

    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 0 int 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; }
    转载请注明原文地址: https://ju.6miu.com/read-40032.html

    最新回复(0)