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：

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 sort O(n).

Space: O(1).

AC Java:

1 public class Solution {
2     public int minMoves2(int[] nums) {
3         int moves = 0;
4         int median = findK(nums, nums.length/2, 0, nums.length-1);
5         for(int num : nums){
6             moves += Math.abs(num-median);
7         }
8         return moves;
9     }
10
11     private int findK(int [] nums, int k, int start, int end){
12         if(start >= end){
13             return nums[start];
14         }
15         int m = partition(nums, start, end);
16         if(m == k){
17             return nums[m];
18         }else if(m < k){
19             return findK(nums, k, m+1, end);
20         }else{
21             return findK(nums, k, start, m-1);
22         }
23     }
24
25     private int partition(int [] nums, int start, int end){
26         int pivot = nums[start];
27         int m = start;
28         int n = start + 1;
29         while(n <= end){
30             if(nums
< pivot){
31                 swap(nums, ++m, n);
32             }
33             n++;
34         }
35         swap(nums, start, m);
36         return m;
37     }
38
39     private void swap(int [] nums, int i, int j){
40         int temp = nums[i];
41         nums[i] = nums[j];
42         nums[j] = temp;
43     }
44 }

类似Minimum Moves to Equal Array Elements.