# Find Minimum in Rotated Sorted Array II ### Source - leetcode: [Find Minimum in Rotated Sorted Array II | LeetCode OJ](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/) - lintcode: [(160) Find Minimum in Rotated Sorted Array II](http://www.lintcode.com/en/problem/find-minimum-in-rotated-sorted-array-ii/) ### Problem 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. #### Example Given [4,4,5,6,7,0,1,2] return 0 ### 题解 由于此题输入可能有重复元素，因此在`num[mid] == num[end]`时无法使用二分的方法缩小start或者end的取值范围。此时只能使用递增start/递减end逐步缩小范围。 ### C++ ~~~ class Solution { public: /** * @param num: a rotated sorted array * @return: the minimum number in the array */ int findMin(vector<int> &num) { if (num.empty()) { return -1; } vector<int>::size_type start = 0; vector<int>::size_type end = num.size() - 1; vector<int>::size_type mid; while (start + 1 < end) { mid = start + (end - start) / 2; if (num[mid] > num[end]) { start = mid; } else if (num[mid] < num[end]) { end = mid; } else { --end; } } if (num[start] < num[end]) { return num[start]; } else { return num[end]; } } }; ~~~ ### Java ~~~ public class Solution { /** * @param num: a rotated sorted array * @return: the minimum number in the array */ public int findMin(int[] num) { if (num == null || num.length == 0) return Integer.MIN_VALUE; int lb = 0, ub = num.length - 1; // case1: num[0] < num[num.length - 1] // if (num[lb] < num[ub]) return num[lb]; // case2: num[0] > num[num.length - 1] or num[0] < num[num.length - 1] while (lb + 1 < ub) { int mid = lb + (ub - lb) / 2; if (num[mid] < num[ub]) { ub = mid; } else if (num[mid] > num[ub]){ lb = mid; } else { ub--; } } return Math.min(num[lb], num[ub]); } } ~~~ ### 源码分析 注意`num[mid] > num[ub]`时应递减 ub 或者递增 lb. ### 复杂度分析 最坏情况下 O(n)O(n)O(n), 平均情况下 O(logn)O(\log n)O(logn).