leetcode1365. How Many Numbers Are Smaller Than the Current Numb
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
这速度啥也不奢求了，就是写一个函数，每次返回比当前索引低的数字的数量，然后遍历一下就行。
Runtime: 22 ms, faster than 17.92% of Java online submissions for How Many Numbers Are Smaller Than the Current Number.
Memory Usage: 44.6 MB, less than 29.36% of Java online submissions for How Many Numbers Are Smaller Than the Current Number.
