LeetCode 374:Guess Number Higher or Lower
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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.
猜数字游戏规则如下:
我从1~n中随机选出一个数,你必须猜出来我选的数字是什么。每一次你猜错的时候,我都会告诉你你的猜测值比真实值高还是低。
你可以调用guess(int num)函数,它有三个可能的返回值(-1 1 0)。
-1 : 数字比猜测值更小 1 : 数字比猜测值更大 0 : 猜中例如:
n = 10, 我选择了6.你的函数应当返回6。
简单的二分法题目,但是因为之前对二分法一知半解,导致自己在做二分法的题目时一直出现错误。。。学习还是不能半桶水啊
// Forward declaration of guess API.// @param num, your guess// @return -1 if my number is lower, 1 if my number is higher, otherwise return 0int guess(int num);class Solution {public: int guessNumber(int n) { int low=1,high=n,mid=low+(high-low)/2; int ans=guess(mid); while(ans!=0) { if(ans==1) {//My number is higher low=mid; mid=low+(high+1-low)/2; } if(ans==-1) {//My number is lower high=mid; mid=low+(high-low-1)/2; } ans=guess(mid); } return mid; }};
这里解释一下mid=low+(high+1-low)/2;和mid=low+(high-low-1)/2;两句话。
首先,不看里面的+1和-1的话,原句其实应该是mid=low+(high-low)/2;,为什么不直接用mid=(high+low)/2;呢,这是因为high+low的值有可能超过int类型的上限值(2^31-1),因此这里需要将加法改为减法。
接下来再解释+1和-1。举个例子,假设当n=2时猜测的值为2,那么low=1,mid=1,high=2,如果采用mid=low+(high-low)/2,那么进行一轮循环之后mid仍然是1(mid=1+(2-1)/2=1),这样就导致了一个死循环。因此当猜测值比我们的值大时,需要将上限增大1,以防止mid值的不改变。同理,假设当n=3时猜测的值为1,那么第一轮low=1,mid=2,high=3,循环一轮之后发现了什么?mid的值依旧是2,这就导致了另一个死循环。因此,这里我们需要将下限减小1。至于为什么会出这种错误,这是因为int会截断小数所导致的,有兴趣的可以自己推一下。
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