[LeetCode] 462. Minimum Moves to Equal Array Elements II

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]

最少移动次数使数组元素相等II。

题意跟版本一很接近，但是这次并不是将所有的数字都往最大值靠拢，这个题考的是问最少操作多少次可以使所有数字相同，而且这个题的操作次数是针对每一个数字而言的，也就是说，每一个数字动一次，就是一次操作。

思路是找到数组的中位数，然后每个数字和中位数的差值就是操作次数。最后要求的结果就是每个数字的操作次数的累加和。既然是求中位数，比较直观的思路就是用Arrays.sort()先排序，找到数组中间的数字，或者是用快速排序找。就这道题的test case而言，前者更快。

时间O(nlogn), worse case is O(n^2)

空间O(1)

Java函数排序

class Solution {
    public int minMoves2(int[] nums) {
        Arrays.sort(nums);
        int res = 0;
        int mid = nums[nums.length / 2];
        for (int num : nums) {
            res += Math.abs(num - mid);
        }
        return res;
    }
}

快速排序

class Solution {
    public int minMoves2(int[] nums) {
        int res = 0;
        int median = findKthLargest(nums, nums.length / 2 + 1);
        for (int num : nums) {
            res += Math.abs(median - num);
        }
        return res;
    }

    private int findKthLargest(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int left = 0;
        int right = nums.length - 1;
        while (true) {
            int pos = partition(nums, left, right);
            if (pos + 1 == k) {
                return nums[pos];
            } else if (pos + 1 > k) {
                right = pos - 1;
            } else {
                left = pos + 1;
            }
        }
    }

    private int partition(int[] nums, int left, int right) {
        int pivot = nums[left];
        int l = left + 1;
        int r = right;
        while (l <= r) {
            if (nums[l] < pivot && nums[r] > pivot) {
                swap(nums, l++, r--);
            }
            if (nums[l] >= pivot) {
                l++;
            }
            if (nums[r] <= pivot) {
                r--;
            }
        }
        swap(nums, left, r);
        return r;
    }

    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

LeetCode 题目总结