Count of Smaller Numbers After Self

题目地址：

https://leetcode.com/problems/count-of-smaller-numbers-after-self/description/

题目：

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Given nums = [5, 2, 6, 1]

To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.

Return the array [2, 1, 1, 0].

解题思路：

merge sort的思想。

代码：

int[] count;
public List<Integer> countSmaller(int[] nums) {
    List<Integer> res = new ArrayList<Integer>();

    count = new int[nums.length];
    int[] indexes = new int[nums.length];
    for(int i = 0; i < nums.length; i++){
        indexes[i] = i;
    }
    mergesort(nums, indexes, 0, nums.length - 1);
    for(int i = 0; i < count.length; i++){
        res.add(count[i]);
    }
    return res;
}
private void mergesort(int[] nums, int[] indexes, int start, int end){
    if(end <= start){
        return;
    }
    int mid = (start + end) / 2;
    mergesort(nums, indexes, start, mid);
    mergesort(nums, indexes, mid + 1, end);

    merge(nums, indexes, start, end);
}
private void merge(int[] nums, int[] indexes, int start, int end){
    int mid = (start + end) / 2;
    int left_index = start;
    int right_index = mid+1;
    int rightcount = 0;
    int[] new_indexes = new int[end - start + 1];

    int sort_index = 0;
    while(left_index <= mid && right_index <= end){
        if(nums[indexes[right_index]] < nums[indexes[left_index]]){
            new_indexes[sort_index] = indexes[right_index];
            rightcount++;
            right_index++;
        }else{
            new_indexes[sort_index] = indexes[left_index];
            count[indexes[left_index]] += rightcount;
            left_index++;
        }
        sort_index++;
    }
    while(left_index <= mid){
        new_indexes[sort_index] = indexes[left_index];
        count[indexes[left_index]] += rightcount;
        left_index++;
        sort_index++;
    }
    while(right_index <= end){
        new_indexes[sort_index++] = indexes[right_index++];
    }
    for(int i = start; i <= end; i++){
        indexes[i] = new_indexes[i - start];
    }
}