Leetcode: Find Minimum in Rotated Sorted Array

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.

这道题跟Search in Rotated Sorted Array这道题很像，区别在于这道题不是找一个目标值，而是找最小值。所以与那道题的思路类似又有区别。类似在于同样是用binary search找目标区域，通过左边界和中间的大小关系来得到左边或者右边有序。区别在于这一次迭代不会存在找到target就终止的情况，iteration要走到 l > r 的情况才停止。所以这一次思路是确定中点哪一边有序之后，那一边的最小值就是最左边的元素。用这个元素尝试更新全局维护的最小值，然后迭代另一边。这样子每次可以截掉一半元素，所以最后复杂度等价于一个二分查找，是O(logn)，空间上只有四个变量维护二分和结果，所以是O(1)。

 1 public class Solution {

 2     public int findMin(int[] num) {

 3         int l = 0;

 4         int r = num.length - 1;

 5         int min = Integer.MAX_VALUE;

 6         while (l <= r) {

 7             int m = (l + r) / 2;

 8             if (num[m] < num[r]) { //[m, r] is sorted

 9                 if (num[m] < min) {

10                     min = num[m];

11                 }

12                 r = m - 1; // search [l, m-1] for next step

13             }

14             else { //[l, m] is sorted

15                 if (num[l] < min) {

16                     min = num[l];

17                 }

18                 l = m + 1; // search [m+1, r] for next step

19             }

20         }

21         return min;

22     }

23 }