Find Minimum in Rotated Sorted Array II

Find Minimum in Rotated Sorted Array II

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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.

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问题

“选择有序数组最小值”问题延伸：允许数组中有重复值。

这会影响时间复杂度吗？如何影响？为什么？

假设一个有序数组在某个未知中心旋转（如0 1 2 4 5 6 7可能变成4 5 6 7 0 1 2），找到数组中最小的元素。

数组中可能包含重复的元素。

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思路

解题思路课参照Find Minimum in Rotated Sorted Array，当允许重复值时，不同之处是：
1. left == right时，无法判断是三种情况中的哪种，此时采用遍历方式。
2. left != right时，当left <= mid时，最小值在右边；当mid <= right时，最小值在左边。

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代码（Java）

ublic class Solution {    public int findMin(int[] nums) {        return binarySearch(nums,0,nums.length-1);    }    public int binarySearch(int[] nums, int left , int right) {        int length = right - left;        if (length == 0) return nums[left];        if (length == 1) return (nums[left] > nums[right] ? nums[right] : nums[left]);        else {            if (nums[left] < nums[right]){                return nums[left];            } else if(nums[left] > nums[right]){                int mid = (left + right)/2;                if (nums[left] < nums[right]){                    return nums[left];                } else if (nums[0] <= nums[mid] ){                    return binarySearch(nums,mid,right);                } else {                    return binarySearch(nums,left,mid);                }            } else {                for(int i = 1;i<nums.length;i++){                    if(nums[i] < nums[i-1]){                        return nums[i];                    }                }                return nums[0];            }        }    }}

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运行结果