Leetcode374. Guess Number Higher or Lower

题目描述：

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.

题目含义：

系统在1到n之间随机选择一个数字，让我们猜这个数字，我们猜的比真实的数大时返回1，小于真实的数时返回-1，相等返回0.

解题思路：

这道题属于典型的二分法，对于1到n这n个数，都是进行有序的排列的，那么我们可以直接套用之前二分的公式进行实现。

具体实现：

/* The guess API is defined in the parent class GuessGame.
    @param num, your guess
    @return -1 if my number is lower, 1 if my number is higher, otherwise return 0

    int guess(int num); */

    public class Solution extends GuessGame {
        public int guessNumber(int n) {
            int start = 1;
            int end = n;
            while (start + 1 < end) {
                int mid = start + (end - start)/2;
                if (guess(mid) == 0) {
                    return mid;
                } else if (guess(mid) == -1) {
                    end = mid;
                } else if (guess(mid) == 1) {
                    start = mid;
                }
            }
            if (guess(start) == 0) {
                return start;
            }
            if (guess(end) == 0) {
                return end;
            }
            return -1;
            }
        }
    }   

时间复杂度为O(logn)，空间复杂度O(1)。