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.
class Solution {
public:
int findMin(vector<int> &num) {
int left = 0, right = num.size()-1;
while(left < right)
{
int mid = left + (right - left) / 2;
if(num[mid] < num[right])
right = mid;
else if(num[mid] > num[right])
left = mid + 1;
}
return num[left];
}
};

本文介绍了一种高效查找旋转排序数组中最小元素的方法。通过使用二分查找算法,可以在O(log n)的时间复杂度内找到数组中的最小值,假设数组中没有重复元素。
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