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 <= 5000 <= 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。
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 }
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