LeetCode 154 Find Minimum in Rotated Sorted Array II (二分 或 分治)

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

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.

题目链接：https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/

题目分析：和Find Minimum in Rotated Sorted Array比就是多了元素可重复的条件，在原来基础上改

public class Solution {    public int findMin(int[] nums) {        int l = 0, r = nums.length - 1, mid;        while (l < r) {            mid = (l + r) >> 1;            if (nums[mid] == nums[r]) {                r --;            }            else if (nums[mid] > nums[r]) {                l = mid + 1;            }            else {                r = mid;            }        }        return nums[l];    }}

其实就是找最小值。。。直接分治，也是logn的复杂度

public class Solution {        int DFS (int[] nums, int l, int r) {        if (l == r) {            return nums[l];        }        if (l == r - 1) {            return Math.min(nums[l], nums[r]);        }        int mid = (l + r) >> 1;        return Math.min(DFS(nums, l, mid), DFS(nums, mid + 1, r));    }        public int findMin(int[] nums) {        return DFS(nums, 0, nums.length - 1);    }}