Given the arraynums, for eachnums[i]find out how many numbers in the array are smaller than it. That is, for eachnums[i]you have to count the number of validj'ssuch thatj != iandnums[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

有多少小於當前數字的數字。題目即是題意，對於每一個數字nums[i]，請你找出陣列中到底有多少個數字小於他。

暴力解很簡單，對於每一個數字nums[i]，我們再次掃描陣列，看看有多少個數字小於他。複雜度是O(n^2)。介於資料量不大，其實暴力解還是可以通過的。

最優解的思路是counting sort/bucket sort。因為資料的限制裡面有這麼一條，這個條件可以幫助我們確定bucket的數量。

0 <= nums[i] <= 100

所以我們可以建立101個桶子，然後遍歷input陣列，計算這101個數字每個數字的frequency。之後從0開始，往後累加每個數字的frequency。最後返回結果的時候，看一下對於每個數字nums[i]，他在桶子對應座標下的值是多少。

時間O(n)

空間O(n)

Java實現

 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 }

相關題目

315.Count of Smaller Numbers After Self

1365. How Many Numbers Are Smaller Than the Current Number

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