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.
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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]
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Constraints:
2 <= nums.length <= 5000 <= nums[i] <= 100
這速度啥也不奢求了，就是寫一個函數(shù)，每次返回比當(dāng)前索引低的數(shù)字的數(shù)量，然后遍歷一下就行。
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.
