标签：
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.
分析：
对不包含重复元素的数组A[0, ... , n - 1], 如果A[mid] < A[high], 说明最小值在A[low, mid]之间。
否则，在A[mid + 1, high]之间。
1 int findMin(vector<int> &num) 2 { 3 int low = 0, high = num.size() - 1, mid = 0; 4 5 while (low < high) 6 { 7 mid = (low + high) / 2; 8 if (num[mid] < num[high]) 9 high = mid; 10 else 11 low = mid + 1; 12 } 13 14 return num[low]; 15 }
如果A包含重复元素，需要加入判断A[mid] == A[high], 此时不能使用二分，如 [1, 1, 0, 1]和[0, 1, 1, 1]， 最小元素
可以在A[mid, high]， 也可能在A[low, mid]， 此时可以high--, 减少区间范围。
1 int findMin(vector<int> &num) 2 { 3 int low = 0, high = num.size() - 1, mid = 0; 4 while (low < high) 5 { 6 mid = (low + high) / 2; 7 if (num[mid] == num[high]) 8 high--; 9 else if (num[mid] < num[high]) 10 high = mid; 11 else 12 low = mid + 1; 13 } 14 15 return num[low]; 16 }
leetcode 154/153. Find Minimum in Rotated Sorted Array && II
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原文地址：http://www.cnblogs.com/ym65536/p/4214573.html
