[Leetcode] 153. Find Minimum in Rotated Sorted Array
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题目:
Suppose an array sorted in ascending order 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.
思路:
这道题目用O(n)的时间复杂度肯定是不允许的,其考点就在于如何变种一下二分查找。每次找到中点,并和最右边的数比较:如果中点大于右边的数,说明最小值一定在右边,则修改左边界;否则说明最小值在左边,则修改左边界。注意一下特殊情况的返回即可。时间复杂度O(logn),空间复杂度O(1)。
PS:明明可以用迭代实现,我却傻不拉几的用了递归实现,导致运行速度大增。这就是我和明明的区别^_^。
代码:
class Solution {public: int findMin(vector<int>& nums) { if (nums.size() == 0) { return 0; } int left = 0, right = nums.size()-1; while(left <= right) { int mid = left + (right - left) / 2; if(nums[mid] == nums[right]) { return nums[mid]; } if(nums[mid] > nums[right]) { left = mid + 1; } else { right = mid; } } return nums[left]; }};
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