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.
The array may contain duplicates.
分析：此题的难点在于duplicate的处理，最坏情况（全部是相同数）时间复杂度是O(n)。此题在实现过程中，返回条件跟Find Minimum in Rotated Sorted Array I不同，优点是不会漏掉一些特殊情况且易于理解。代码如下：
class Solution { public: int findMin(vector<int> &num) { int n = num.size(); if(n == 1) return num[0]; if(n == 2) return min(num[0], num[1]); if(num[0] < num[n-1]) return num[0]; int l = 0, r = n - 1; while(l <= r){ if(l == r) return num[l]; if(l == r-1) return min(num[l], num[r]); int mid = (l + r)/2; if(num[mid] > num[0]) l = mid + 1; else if(num[mid] < num[0]) r = mid; else if(num[0] > num[n-1]) l = mid + 1; else {//deal with duplicates if(mid > n-1-mid){ int tmp = mid; while(tmp < n-1 && num[tmp] == num[tmp+1]) tmp++; if(tmp == n-1) r = mid; else l = tmp + 1; }else { int tmp = mid; while(tmp > 0 && num[tmp-1] == num[tmp]) tmp--; if(tmp == 0) l = mid; else r = tmp - 1; } } } } };
Find Minimum in Rotated Sorted Array II
原文：http://www.cnblogs.com/Kai-Xing/p/4059785.html