lintcode Guess Number Game(Java)

最新推荐文章于 2021-04-15 23:32:46 发布

原创最新推荐文章于 2021-04-15 23:32:46 发布 · 1.5k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法

算法专栏收录该内容

19 篇文章

订阅专栏

本文介绍了一种经典的猜数字游戏的高效解决方案——二分查找法。通过定义预先设定的API来判断每次猜测的数字与目标数字的关系，最终找到正确答案。文章详细解释了如何避免整数溢出的问题，并给出了具体的实现代码。

1.题目

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):

详解：这就是我们常玩的猜数字游戏，例如从1-100，取数字20，每次猜一个（提供了API guess(int num)），等于20，返回0,大于20返回-1，小于20返回1，直到返回0为止。

样例
n = 10, I pick 4 (but you don’t know)
Return 4. Correct !

2.解法

其实就是二分法解题，关键的一步就是计算中间节点。
int middle = (start + end)/2,有可能导致溢出，下面的方法更好
int middle = start + (end - start)/2;//中间节点的算法

/* 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 {
    /**
     * @param n an integer
     * @return the number you guess
     */
    public int guessNumber(int n) {
        // Write your code here
         int start = 0;
        int end = n;

        while (start < end){
            int middle = start + (end - start)/2;//中间节点的算法
            int result = guess(middle);
            if (result == -1){
                end = middle - 1;
            }else if (result == 1){
                start = middle + 1;
            }else if (result == 0){
                return middle;
            }
        }
           return start;
    }
}