Description
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
分析
题目的意思是：给你一个数组，统计比当前数小的数的个数，首先用了pos字典把所有的坐标存放了下来，然后对原来的数据进行了排序，然后遍历排序的数组就能很容易得到结果了。看别人的代码发现有比这更好的思路，首先用sorted排序，用字典存放比当前数小的个数，最后把统计结果放回到对应的位置就行了。
代码
class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
s=[]
res=[0]*len(nums)
pos=collections.defaultdict(list)
for i in range(len(nums)):
pos[nums[i]].append(i)
nums.sort()
for num in nums:
cnt=len(s)
while(len(pos[num])>0):
res[pos[num].pop()]=cnt
s.append(num)
return res
参考文献
[LeetCode] Python 3 -> 88.46% faster. 56ms time. Explanation added
