_315_CountofSmallerNumbersAfterSelf.java

package leetcode_1To300;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
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 * - 版权所有，转发请注明出处
 */
public class _315_CountofSmallerNumbersAfterSelf {
    /**
     * 315. Count of Smaller Numbers After Self
     * 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].

     [5, 2, 6, 1]

     int[] Arrays.asList()
     0 1 1 2

     time : O(n^2)
     space : O(n)
     */
    public List<Integer> countSmaller(int[] nums) {
        Integer[] res = new Integer[nums.length];
        List<Integer> list = new ArrayList<>();
        for (int i = nums.length - 1; i >= 0; i--) {
            int index = findIndex(list, nums[i]);
            res[i] = index;
            list.add(index, nums[i]);
        }
        return Arrays.asList(res);
    }

    private int findIndex(List<Integer> list, int target) {
        if (list.size() == 0) return 0;
        int start = 0;
        int end = list.size() - 1;
        if (list.get(end) < target) return end + 1;
        if (list.get(start) >= target) return 0;
        while (start + 1 < end) {
            int mid = (end - start) / 2 + start;
            if (list.get(mid) < target) {
                start = mid + 1;
            } else {
                end = mid;
            }
        }
        if (list.get(start) >= target) return start;
        return end;
    }
}