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Title：
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>& nums) { int left = 0; int right = nums.size()-1; while (left < right && nums[left] > nums[right]){ int mid = (left + right) / 2; if (nums[mid] < nums[right]) right = mid; else left = mid + 1; } return nums[left]; } };
II
The array may contain duplicates.
只需要修改while语句，并添加一个if语句。 [1,0,1,1,1]
class Solution { public: int findMin(vector<int>& nums) { int left = 0; int right = nums.size()-1; while (left < right && nums[left] >= nums[right]){ int mid = (left + right) / 2; if (nums[mid] < nums[right]) right = mid; else if (nums[mid] > nums[right]) left = mid + 1; else left += 1; } return nums[left]; } };
LeetCode: Find Minimum in Rotated Sorted Array
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原文地址：http://www.cnblogs.com/yxzfscg/p/4530399.html
