LeetCode 462. Minimum Moves to Equal Array Elements II-优快云博客

本文链接：https://blog.youkuaiyun.com/David_DLUT/article/details/53932936

本文介绍了一种算法，用于计算使整数数组所有元素相等所需的最小移动次数。通过找到中位数并计算各元素与中位数的距离之和来解决问题。

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Given a non-empty integer array, find the minimum number of moves required to make all array elements equal, where a move is incrementing a selected element by 1 or decrementing a selected element by 1.

You may assume the array's length is at most 10,000.

Example:

Input:
[1,2,3]

Output:
2

Explanation:
Only two moves are needed (remember each move increments or decrements one element):

[1,2,3]  =>  [2,2,3]  =>  [2,2,2]

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class Solution {
public:
    int minMoves2(vector<int>& nums) {
        int result = 0;
        int n = nums.size();
        int median = medianVal(nums);
        for (int i = 0; i < n; ++i) {
            result += abs(nums[i] - median);
        }
        return result;
    }
private:
    inline void swap(int& a, int& b) {
        int temp = a;
        a = b;
        b = temp;
    }
    int medianVal(vector<int>& nums) {
        int n = nums.size();
        int left = 0, right = n - 1;
        int mid = n / 2;
        while (true) {
            int k = left;
            int temp = rand() % (right - left + 1) + left;
            swap(nums[right], nums[temp]);
            for (int i = left; i < right; ++i) {
                if (nums[i] < nums[right]) {
                    swap(nums[i], nums[k++]);
                }
            }
            swap(nums[right], nums[k]);
            if (k == mid) {
                return nums[k];
            }
            else if (k < mid) {
                left = k + 1;
            }
            else {
                right = k - 1;
            }
        }
    }
};