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| // @Title: 单词拆分 II (Word Break II)
// @Author: 15816537946@163.com
// @Date: 2019-09-06 23:23:43
// @Runtime: 4 ms
// @Memory: 2.6 MB
/*
* @lc app=leetcode.cn id=140 lang=golang
*
* [140] 单词拆分 II
*
* https://leetcode-cn.com/problems/word-break-ii/description/
*
* algorithms
* Hard (37.39%)
* Likes: 69
* Dislikes: 0
* Total Accepted: 5.8K
* Total Submissions: 15.6K
* Testcase Example: '"catsanddog"\n["cat","cats","and","sand","dog"]'
*
* 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:
* []
*
*
*/
func wordBreak(s string, wordDict []string) []string {
mp := make(map[string]bool)
for _, v := range wordDict {
mp[v] = true
}
// 1. 动态规划, 判断字符串是否能被拆分
dp := make([]bool,len(s)+1)
dp[0] = true
for i := 1; i<= len(s);i++ {
for j:=0;j<i;j++ {
if dp[j] && mp[s[j:i]] {
dp[i] = true
break
}
}
}
// 2. 判断s是否可以被wordDict里的元素分割
ret := make([]string,0)
if !dp[len(s)] {
return ret
}
// 3. dfs
var DFS = func(string, []string) {}
DFS = func(ns string, r []string) {
if len(ns) == 0 {
ret = append(ret, strings.Join(r," "))
return
}
for i:=1;i<=len(ns);i++ {
if mp[ns[:i]] {
DFS(ns[i:],append(r, ns[:i]))
}
}
}
DFS(s, []string{})
return ret
}
|