462. Minimum Moves to Equal Array Elements II

Given an integer array nums of size n, return the minimum number of moves required to make all array elements equal.

In one move, you can increment or decrement an element of the array by 1.

Example 1:

Input: nums = [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]

Example 2:

Input: nums = [1,10,2,9]
Output: 16

Constraints:

  • n == nums.length
  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109
class Solution {
    public int minMoves2(int[] nums) {
         Arrays.sort(nums);
        int t = nums[nums.length / 2];
        int res = 0;
        for(int i : nums) res += Math.abs(t - i);
        return res;
    }
}

找中间

class Solution {
    public int minMoves2(int[] nums) {
         Arrays.sort(nums);
        int t = help(nums, 0, nums.length - 1, nums.length / 2 );
        int res = 0;
        for(int i : nums) res += Math.abs(t - i);
        return res;

    }
    public int help(int[] nums, int low, int high, int k) {
        int i = low - 1;
        int pivot = nums[high];
        
        for(int j = low; j < high; j++){
            if(nums[j] < pivot){
                i++;
                swap(i, j, nums);
            }
        }   
        swap(i+1, high, nums);
        if(i+1 == k) return nums[i+1];
        else if(i + 1 < k) return help(nums, i + 2, high, k);
        else return help(nums, low, i, k);
    }
    public void swap(int i, int j, int[] nums){
        int a = nums[i] + nums[j];
        nums[j] = a - nums[j];
        nums[i] = a - nums[j];
    }
}

用快select还慢了。。

原文地址:https://www.cnblogs.com/wentiliangkaihua/p/14788333.html