(前缀树) leetcode 1365. 有多少小于当前数字的数字 472. 连接词

1365. 有多少小于当前数字的数字

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        num_count = [0]*101
        for num in nums:
            num_count[num]+=1
        # print(num_count)
        start = 0
        for i,v in enumerate(num_count):
            num_count[i]=start
            start+=v
        # print(num_count)
        result = [0]*len(nums)
        for i,n in enumerate(nums):
            result[i] = num_count[n]
        return result

472. 连接词

from functools import lru_cache
class Solution:
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        Tree = {
     'children':[None]*26,'is_word':False}
        result = []
        for word in words:
            p = Tree
            for i,c in enumerate(word):
                if p['children'][ord(c)-ord('a')]:
                    if i==len(word)-1:
                        p['children'][ord(c)-ord('a')]['is_word']=True
                    p = p['children'][ord(c)-ord('a')]
                else:
                    tmp = {
     'children':[None]*26,'is_word':False}
                    p['children'][ord(c)-ord('a')] = tmp
                    if i==len(word)-1:
                        tmp['is_word']=True
                    p = tmp

        @lru_cache(None)
        def dfs(index,count,word):
            p = Tree
            for i in range(index,len(word)):
                c = word[i]
                # print(word,c)
                if not p['children'][ord(c)-ord('a')]:
                    return False
                if p['children'][ord(c)-ord('a')]['is_word']:   
                    if i==len(word)-1 and count>=1:
                        return True
                    if dfs(i+1,count+1,word) :
                        return True
                p = p['children'][ord(c)-ord('a')]
        for word in words:
            if dfs(0,0,word):
                result+=[word]
        return result