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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]; }
Find Minimum in Rotated Sorted Array——LeetCode
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原文地址：http://www.cnblogs.com/aboutblank/p/4447068.html
