leetcode 140. 单词拆分 II

题目描述

给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,在字符串中增加空格来构建一个句子,使得句子中所有的单词都在词典中。返回所有这些可能的句子。

说明:

  • 分隔时可以重复使用字典中的单词。
  • 你可以假设字典中没有重复的单词。

示例 1:

输入:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
输出:
[
  "cats and dog",
  "cat sand dog"
]
示例 2:
输入:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
输出:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
解释: 注意你可以重复使用字典中的单词。
示例 3:
输入:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
输出:
[]

题解

记忆化回溯,hash[s] 保存相应字符串 s 对应的单词组合

示例代码(go)

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func wordBreak(s string, wordDict []string) []string {
hash := make(map[string][]string)
wordHash := make(map[string]bool)
for _, word := range wordDict {
wordHash[word] = true
}
return dfs(s, wordDict, hash, wordHash)
}

func dfs(s string, wordDict []string, hash map[string][]string, wordHash map[string]bool) []string {
if _, ok := hash[s]; ok {
return hash[s]
}
res := make([]string, 0)
n := len(s)
if n == 0 {
res = append(res, "")
return res
}
for i := 0; i < n; i++ {
if wordHash[s[:i+1]] {
sublist := dfs(s[i+1:], wordDict, hash, wordHash)
for _, sub := range sublist {
space := " "
if sub == "" {
space = ""
}
res = append(res, s[:i+1]+space+sub)
}
}
}
hash[s] = res
return res
}