154. Find Minimum in Rotated Sorted Array II

最新推荐文章于 2025-08-24 11:06:10 发布

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最新推荐文章于 2025-08-24 11:06:10 发布

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文章标签： c/c++ 开发工具 runtime

原文链接：http://www.cnblogs.com/ruruozhenhao/p/9890144.html

本文探讨了在一个未知旋转点的升序数组中查找最小元素的问题，特别关注数组可能包含重复元素的情况。通过分析，提供了一种有效的算法解决方案，并讨论了其运行时复杂性的影响。

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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.

Example 1:

Input: [1,3,5]
Output: 1

Example 2:

Input: [2,2,2,0,1]
Output: 0

Note:

This is a follow up problem to Find Minimum in Rotated Sorted Array.
Would allow duplicates affect the run-time complexity? How and why?

Approach #1:

class Solution {
public:
    int findMin(vector<int>& nums) {
        int len = nums.size();
        int l = 0, r = len - 1;
        while (l <= r) {
            while (nums[l] == nums[l+1] && r > l) l++;
            while (nums[r] == nums[r-1] && r > l) r--;
            int m = l + (r - l) / 2;
            if (l == r) return nums[l];
            if (nums[r] > nums[l])
                return nums[l];
            if (nums[m] > nums[m+1]) return nums[m+1];
            if (nums[m] < nums[m-1]) return nums[m];
            if (nums[m] > nums[l])
                l = m + 1;
            else 
                r = m - 1;
        }
        return -1;
    }
};

Runtime: 4 ms, faster than 98.95% of C++ online submissions for Find Minimum in Rotated Sorted Array II.

Analysis:

this problem have the same idea with `154. Find Minimum in Rotated Sorted Array I`, the only different is when we meet the same element, we should let l++ or r-- in the legal range.

转载于:https://www.cnblogs.com/ruruozhenhao/p/9890144.html