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.
题目大意:给定一个有序数组,是经过旋转过的,不知道在哪儿旋转的,求数组最小元素。
解题思路:直接遍历可以,太low。二分搜索吧,有序数组旋转有个特征,旋转后面的元素比旋转点前面的所有元素都小。
Talk is cheap>>
public int findMin(int[] nums) { if (nums == null || nums.length == 0) { return 0; } int low = 0; int high = nums.length - 1; int mid = 0; while (low <= high) { mid = (low + high) >> 1; int left = mid > 0 ? mid - 1 : 0; int right = mid < nums.length - 1 ? mid + 1 : nums.length - 1; if (nums[mid] <= nums[left] && nums[mid] <= nums[right]) { break; } if (nums[mid] >= nums[high]) { low = mid + 1; } else { high = mid - 1; } } return nums[mid]; }
时间: 2024-10-10 05:35:53