字符串算法——查找有序数组旋转后最小值(有重复元素)(Find Minimum in Rotated Sorted Array II)

问题:
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 an array sorted in ascending order 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.
思路一：同无重复元素数组一样，采用最简单的直接遍历，依次比较，不过这里时间复杂度不满足条件。

class Solution {    public int findMin(int[] nums) {        int len = nums.length;        int minIndex = 0;        for(int i = 1;i<len;i++){            if(nums[minIndex]>nums[i]){                minIndex = i;            }        }        return nums[minIndex]；    }}

思路二：同无重复元素数组一样，可以采用二分查找的算法，但由于存在重复元素，所以无法确定最小值是落在中间索引的哪一侧，对两侧分别查找，比较两侧查找的最小值。

class Solution {    public int findMin(int[] nums) {        int len = nums.length;        return dfs(nums,0,len-1);    }    private int dfs(int[]nums,int min,int max){        if(min==max){            return nums[min];        }        int mid = (min+max)/2;//中间索引值        int left = Integer.MAX_VALUE;        int right = Integer.MAX_VALUE;        if(nums[mid]>=nums[max]){            left = dfs(nums,mid+1,max);//可能落在右侧        }        if(nums[mid]<=nums[max]){//可能落在左侧            right = dfs(nums,min,mid);        }        return Math.min(left,right);//比较两侧查找出的值    }}