leetcode[1365] How Many Numbers Are Smaller Than the Current Number Python3实现（排序，哈希记录排名）

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#leetcode #数据结构 #python #哈希 #快速排序

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本文介绍了一种高效算法，用于解决LeetCode题目中关于统计数组中比当前元素小的元素数量的问题。通过使用排序和哈希表，算法能在O(n log n)的时间复杂度内完成任务。

# 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 v
# alid 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 
#  
#  Related Topics 数组 哈希表 
#  👍 91 👎 0

# leetcode submit region begin(Prohibit modification and deletion)
class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        numsSorted = sorted(nums)
        rankList = [0] * 101
        rank = 0
        for i in range(1, len(numsSorted)):
            if numsSorted[i]!=numsSorted[i-1]: rank = i
            rankList[numsSorted[i]] = rank
        ret = []
        for num in nums:
            ret.append(rankList[num])
        return ret
# leetcode submit region end(Prohibit modification and deletion)

求比某个元素小的元素个数，只需求出该元素在数组中的排序次序。

由于0 <= nums[i] <= 100，可用rankList(101)模拟哈希表记录元素的排名位次。

最后遍历一遍原始数组，将相应位次添加到ret中返回即可。

拿空间换时间最成功的一次：

解答成功:
执行耗时:36 ms,击败了99.27% 的Python3用户
内存消耗:13.7 MB,击败了5.09% 的Python3用户