[LeetCode] 1365. How Many Numbers Are Smaller Than the Current Number

时间：2020-07-11 09:17:31 阅读：48 评论：0 收藏：0 [点我收藏+]

标签：exist ret str mic array enc png const tput

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

有多少小于当前数字的数字。题目即是题意，对于每一个数字nums[i]，请你找出数组中到底有多少个数字小于他。

暴力解很简单，对于每一个数字nums[i]，我们再次扫描数组，看看有多少个数字小于他。复杂度是O(n^2)。介于数据量不大，其实暴力解还是可以通过的。

技术图片

最优解的思路是counting sort/bucket sort。因为数据的限制里面有这么一条，这个条件可以帮助我们确定bucket的数量。

0 <= nums[i] <= 100

所以我们可以创建101个桶子，然后遍历input数组，计算这101个数字每个数字的frequency。之后从0开始，往后累加每个数字的frequency。最后返回结果的时候，看一下对于每个数字nums[i]，他在桶子对应坐标下的值是多少。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public int[] smallerNumbersThanCurrent(int[] nums) {
 3         int[] freq = new int[101];
 4         // count the frequency
 5         for (int num : nums) {
 6             freq[num]++;
 7         }
 8         // sum up all the frequencies
 9         for (int i = 1; i < freq.length; i++) {
10             freq[i] += freq[i - 1];
11         }
12         // output
13         int[] res = new int[nums.length];
14         for (int i = 0; i < res.length; i++) {
15             if (nums[i] > 0) {
16                 res[i] = freq[nums[i] - 1];
17             }
18         }
19         return res;
20     }
21 }