• [leetcode] 278. First Bad Version


    You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

    Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.

    You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

    我的代码如下:

     1 // Forward declaration of isBadVersion API.
     2 bool isBadVersion(int version);
     3 
     4 class Solution {
     5 public:
     6     int firstBadVersion(int n) {
     7         if (isBadVersion(1)) return 1;
     8         int left = 1, right = n, mid = (1 + n) / 2;
     9         while (left < right) {
    10             mid = left + (right - left) / 2;
    11             if (isBadVersion(mid)) right = mid;
    12             else left = mid + 1;
    13         }
    14         return left;
    15     }
    16 };

    一开始我一直把第10行写为

    mid = (left + right) / 2;

    但是一直超时,后来搜了一下发现 因为(left + right) 有可能会溢出导致不断循环。

    但是有一个困惑并没有找到原因:

    把第10行换到else下边就会超时。。。希望能有人解惑。

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  • 原文地址:https://www.cnblogs.com/zmj97/p/7502635.html
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