# Find Minimum in Rotated Sorted Array ### Source - leetcode: [Find Minimum in Rotated Sorted Array | LeetCode OJ](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) - lintcode: [(159) Find Minimum in Rotated Sorted Array](http://www.lintcode.com/en/problem/find-minimum-in-rotated-sorted-array/) ### 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. #### Example Given `[4, 5, 6, 7, 0, 1, 2]` return `0` #### Note You may assume no duplicate exists in the array. ### 题解 如前节所述，对于旋转数组的分析可使用画图的方法，如下图所示，升序数组经旋转后可能为如下两种形式。  最小值可能在上图中的两种位置出现，如果仍然使用数组首部元素作为target去比较，则需要考虑图中右侧情况。**使用逆向思维分析，如果使用数组尾部元素分析，则无需图中右侧的特殊情况。**不过考虑在内的话也算是一种优化。 ### 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]) { end = mid; } else { start = mid; } } 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 { lb = mid; } } return Math.min(num[lb], num[ub]); } } ~~~ ### 源码分析 仅需注意使用`num[end]`(使用 num[lb]不是那么直观)作为判断依据即可，由于题中已给无重复数组的条件，故无需处理`num[mid] == num[end]`特殊条件。 ### 复杂度分析 由于无重复元素，平均情况下复杂度为 O(logn)O(\log n)O(logn).