Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array.
题目的意思是想在头尾连在一起是一个排好序的数组里面找出最小的那个数
感觉很简单,于是有了下面的解法
代码如下:
class Solution { public: int findMin(vector<int>& nums) { if (nums.empty()) return -1; int min = nums[0]; for (int i = 0; i < nums.size(); i++) { if (nums[i] < min) return nums[i]; } return min; } };
Custom Testcase:[4,5,6,7,0]
result: 0
时间: 2024-12-28 17:29:28