315. 计算右侧小于当前元素的个数

给你一个整数数组 nums ，按要求返回一个新数组 counts 。数组 counts 有该性质： counts[i] 的值是  nums[i] 右侧小于 nums[i] 的元素的数量。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/count-of-smaller-numbers-after-self
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

class Solution {

    private int[] ans;

    /**
     * 当前 i 位置的数原来在 index[i]
     */
    private int[] index;

    private int[] helper;

    private int[] indexHelper;

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

        int mid = (l + r) >> 1;

        mergeSort(nums, l, mid);
        mergeSort(nums, mid + 1, r);

        int p1 = l, p2 = mid + 1, p = l;
        while (p1 <= mid && p2 <= r) {
            if (nums[p1] <= nums[p2]) {
                indexHelper[p] = index[p2];
                helper[p++] = nums[p2++];
            } else {
                ans[index[p1]] += (r - p2 + 1);
                indexHelper[p] = index[p1];
                helper[p++] = nums[p1++];
            }
        }

        while (p1 <= mid) {
            indexHelper[p] = index[p1];
            helper[p++] = nums[p1++];
        }

        while (p2 <= r) {
            indexHelper[p] = index[p2];
            helper[p++] = nums[p2++];
        }

        for (int i = l; i <= r; ++i) {
            index[i] = indexHelper[i];
        }

        System.arraycopy(helper, l, nums, l, r - l + 1);
    }

    public List<Integer> countSmaller(int[] nums) {
        if (nums == null || nums.length == 0) {
            return Collections.emptyList();
        }

        this.ans = new int[nums.length];
        this.index = new int[nums.length];
        this.helper = new int[nums.length];
        this.indexHelper = new int[nums.length];

        for (int i = 0; i < nums.length; ++i) {
            index[i] = i;
        }

        int[] copy = new int[nums.length];
        System.arraycopy(nums, 0, copy, 0, nums.length);

        mergeSort(copy, 0, nums.length - 1);

        List<Integer> ret = new ArrayList<>(ans.length);

        for (int i = 0; i < ans.length; ++i) {
            ret.add(ans[i]);
        }

        return ret;
    }
}

标签：nums,int,个数,315,length,private,ans,new,右侧
来源： https://www.cnblogs.com/tianyiya/p/15662663.html