LeetCode Minimum Moves to Equal Array Elements II

原题链接在这里：https://leetcode.com/problems/minimum-moves-to-equal-array-elements-ii/

题目：

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]

题解：

The median minimizes the sum of absolute deviations. There is an article explaining this.

用quick sort找到median. 类似Kth Largest Element in an Array.

每一个元素对于median差值的绝对值的总和就是最小moves. 这里的median可以理解为第nums.length/2小的element.

Time Complexity: O(n), quick select O(n). n = nums.length.

Space: O(1).

AC Java:

 class Solution {
     public int minMoves2(int[] nums) {
         if(nums == null || nums.length < 2){
             return 0;
         }

         int n = nums.length;
         int median = findK(nums, (n - 1) / 2, 0, n - 1);
         int res = 0;
         for(int num : nums){
             res += Math.abs(num - median);
         }

         return res;
     }

     private int findK(int [] nums, int k, int l, int r){
         if(l >= r){
             return nums[l];
         }

         int m = partition(nums, l, r);
         if(m == k){
             return nums[k];
         }else if(m < k){
             return findK(nums, k, m + 1, r);
         }else{
             return findK(nums, k, l, m - 1);
         }
     }

     private int partition(int [] nums, int l, int r){
         int pivot = nums[l];
         while(l < r){
             while(l < r && nums[r] >= pivot){
                 r--;
             }

             nums[l] = nums[r];

             while(l < r && nums[l] <= pivot){
                 l++;
             }

             nums[r] = nums[l];
         }

         nums[l] = pivot;
         return l;
     }
 }

类似Minimum Moves to Equal Array Elements, Best Meeting Point.