Leetcode_find-minimum-in-rotated-sorted-array-ii（c++ version）-优快云博客

探讨在允许重复元素的旋转排序数组中查找最小值的方法。当数组中存在重复元素时，传统的二分查找法可能退化为线性搜索。文章通过递归方式解决此问题，并提供了一个C++实现示例。

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

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.

思路：

跟上一题类似，但是如果num[0] == num[mid]的话，两个部分都要子递归。这时是 f(n) = 2f((n-1)/2) , 时间复杂度是O(2^logn) = O(n)

故最好时间复杂度是o(logn), 最坏时间复杂度是o(n).

参考代码：

class Solution {
public:
    int findMin(vector<int> &num) {
        if(num.size()==1)
            return num[0];
        else if(num.size()==2)
            return min(num[0], num[1]);
            
        int mid = num.size()/2;
        if(num[0] == num[mid]) {
            vector<int>vec1(num.begin(), num.begin()+mid);
            vector<int>vec2(num.begin()+mid+1, num.end());
            return min(findMin(vec1), findMin(vec2));
        } else if(num[0] < num[mid]) {
            vector<int>vec(num.begin()+mid+1, num.end());
            return min(num[0], findMin(vec));
        } else {
            vector<int>vec(num.begin(), num.begin()+mid+1);
            return findMin(vec);
        }
    }
};