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Leetcode 1048. Longest String Chain

2021/7/21 16:23:34 浏览:

文章作者:Tyan
博客:noahsnail.com  |  CSDN  |  简书

1. Description

Longest String Chain

2. Solution

**解析:**Version 1,先根据字符串长度对数组排序,然后根据长度分到不同的组里,按长度遍历组,如果下一组的字符串长度比当前组多1个,则遍历两组的所有元素,满足条件前辈子串,则下一组子串的字符链长度在当前子串长度的基础上加1,其实就是一个广度优先搜索的过程。Version 2遍历字符串所有长度减1的子串,如果找到,则比较字符链长度,判断是否需要加1,返回最大长度。

  • Version 1
class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        stat ={}
        words.sort(key=len)
        for word in words:
            stat[len(word)] = stat.get(len(word), []) + [word]
        chains = {word: 1 for word in words}
        for k, v in stat.items():
            if k+1 in stat:
                for a in v:
                    for b in stat[k+1]:
                        if chains[b] <= chains[a] and self.predecessor(a, b):
                            chains[b] = chains[a] + 1
        return max(chains.values())


    def predecessor(self, a, b):
        i = 0
        j = 0
        while i < len(a) and j < len(b):
            if a[i] == b[j]:
                i += 1
                j += 1
            else:
                j += 1
            if j - i > 1:
                return False
        return True

Version 2

class Solution:
    def longestStrChain(self, words: List[str]) -> int:
        words.sort(key=len)
        stat = {word: 1 for word in words}
        for word in words:
            for i in range(len(word)):
                pre = word[:i] + word[i+1:]
                if pre in stat:
                    stat[word] = max(stat[word], stat[pre] + 1)
        return max(stat.values())

Reference

  1. https://leetcode.com/problems/longest-string-chain/

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