题目描述

https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/

Given the array nums, for each nums[i] find out how many numbers in the array are 
smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].
Return the answer in an array.

Example 1:

Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation: 
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). 
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1). 
For nums[3]=2 there exist one smaller number than it (1). 
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
Example 2:

Input: nums = [6,5,4,8]
Output: [2,1,0,3]
Example 3:

Input: nums = [7,7,7,7]
Output: [0,0,0,0]
 

Constraints:

2 <= nums.length <= 500
0 <= nums[i] <= 100

博主提交的代码

解法一 简单直接，时间复杂度O(n^2)，空间复杂度O(1)

这个正常人直观都能想到的解法

public int[] smallerNumbersThanCurrent(int[] nums) {
        int[] result = new int[nums.length];
        for(int i = 0; i < nums.length; i++){
            for(int eachInnerInt: nums){
                if( nums[i] > eachInnerInt){
                    result[i]++;
                }
            }
        }
        return result;
    }

解法二，先排序再输出，时间复杂度依赖于你的排序算法

我们采用jdk默认的排序算法来进行排序

Arrays.sort jdk1.8官方解释

public int[] smallerNumbersThanCurrent(int[] nums) {
        // 需要clone的原因是Arrays.sort会改变入参的顺序
        int[] backup = nums.clone();
        // Jdk 1.8的Arrays.sort的官方javadoc中已经写明此方法原理和时间复杂度，为nlogn
        Arrays.sort(backup);
        Map<Integer,Integer> dataIndex = new HashMap<>();
        int ordinal = 0;
        for(int sortedInt: backup){
            if( dataIndex.get(sortedInt) == null){
                dataIndex.put(sortedInt,ordinal);
            }
            ordinal++;
        }
        int[] result = new int[nums.length];
        for(int i = 0;i < nums.length; i++){
            result[i] = dataIndex.get(nums[i]);
        }
        return result;
    }

他人优秀的代码

解法一

这个解法是nums里面的值比较大的时候，并不太适用，但是在这道题的背景下还是可以的

public int[] smallerNumbersThanCurrent(int[] nums) {
        int n = nums.length;

        int[] c1 = new int[101]; // 0 <= nums[i] <= 100
        for (int i : nums) {
            c1[i]++;
        }

        int[] c2 = new int[101];
        c2[0] = c1[0];
        for (int i = 1; i < 101; i++) {
            c2[i] = c1[i] + c2[i - 1];
        }

        int[] ans = new int[n];
        int idx = 0;
        for (int i : nums) {
            ans[idx++] = c2[i] - c1[i];
        }
        return ans;
    }