题目

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

分析

题意：对于数组中的每个数字，求出整个数组中比他小的数组的个数。

最简单的解法就是对于每个数字，都遍历一遍数组挨个比较，这样的话，复杂度显然是n的n次方。
有没有更好的解法？

很容易想到，将数组按升序排序，下标i，就是整个数组中比他小的数组的个数.
但是，对于重复的数据，下标i就不准确了。因此要对其去重。

解答

我写了一个，报错了。应该是遍历出错了，这个算法应该是双层遍历才对，我只写了一个。
排序去重的数组长度是MAX值101（nums[i] <= 100），nums的长度不确定，我不太清楚怎么遍历获取。

class Solution {public int[] smallerNumbersThanCurrent(int[] nums) {int len = nums.length;int[] res = new int[len];int[] afterSD = sortAndDist(nums);int i=0;int j=0;int k=0;while(k<len){if(afterSD[i]==0){i++;continue;}else if(afterSD[i]!=nums[j]){i++;j++;continue;}else {res[k]=i;k++;}}return res;}public int[] sortAndDist(int[] nums){int[] tmp = new int[101];for(int i=0;i<nums.length;i++){tmp[nums[i]]=1;}return tmp;}
}

看看别人写的。
做法类似，但是思路不太一样，
我想的算法还有个漏洞，比如[1,2,3,3,4]，如果对其去重排序，得到的是
[1,2,3,4]
那么比1小的有0个，比2小的有1个，比3小的有2个，到这里都是正确的，
但是比4小的有3个，显然是错误的。
因此去重还得记录重复个数才行。

class Solution {public int[] smallerNumbersThanCurrent(int[] nums) {int[] count = new int[101];int[] res = new int[nums.length];for (int i =0; i < nums.length; i++) {// 这里用的自增1的形式，count的数字表示对应下标的数据的重复个数count[nums[i]]++;}for (int i = 1 ; i <= 100; i++) {// 把前一个数的个数，加到后一个上面，这样，就统计了比当前数小的数据个数// 比如count=[0,1,2,3]，// 意思是比0小的0个，// 比1小的有1个// 比i小的有count[i]个// 累加之后count=[0,1,3,6],显然是正确的。count[i] += count[i-1];    }for (int i = 0; i < nums.length; i++) {// 如果是0，显然比他小的只有0个if (nums[i] == 0)res[i] = 0;else  res[i] = count[nums[i] - 1];}return res;        }
}

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40 How Many Numbers Are Smaller Than the Current Number

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