Word Break

问题描述

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given

s =

"leetcode"

,

dict =

["leet", "code"]

.

Return true because

"leetcode"

can be segmented as

"leet code"

.

解决思路

1. dfs，超时；

2. dp。

使用一个大小为输入字符串长度加1的辅助数组，dp[i]表示S[0, i]字符串是否可以被分割。

双重for循环，时间复杂度为O(n^2).

程序

1. DFS

public class Solution {
public boolean wordBreak(String s, Set<String> wordDict) {
if (s == null || s.length() == 0) {
return true;
}
if (wordDict == null || wordDict.size() == 0) {
return false;
}
return helper(s, wordDict);
}

private boolean helper(String s, Set<String> wordDict) {
if (s.length() == 0) {
return true;
}
boolean flag = false;
for (int i = 0; i < s.length(); i++) {
String word = s.substring(0, i + 1);
if (wordDict.contains(word)) {
flag = helper(s.substring(i + 1), wordDict);
}
}
return flag;
}
}

2. DP

public class Solution {
public boolean wordBreak(String s, Set<String> wordDict) {
if (s == null || s.length() == 0) {
return true;
}
if (wordDict == null || wordDict.size() == 0) {
return false;
}
return helper(s, wordDict);
}

private boolean helper(String s, Set<String> wordDict) {
if (s.length() == 0) {
return true;
}
boolean flag = false;
for (int i = 0; i < s.length(); i++) {
String word = s.substring(0, i + 1);
if (wordDict.contains(word)) {
flag = helper(s.substring(i + 1), wordDict);
}
}
return flag;
}
}

　　

WordBreakII

问题描述

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given

s =

"catsanddog"

,

dict =

["cat", "cats", "and", "sand", "dog"]

.

A solution is

["cats and dog", "cat sand dog"]

.

解决思路

还是动态规划的思路，区别问题I的是，dp[i]代表以i为结尾的，能够break的单词，由于单词数可能有多个，所以是一个list数组。

程序

public class Solution {
public List<String> wordBreak(String s, Set<String> wordDict) {
List<String> res = new ArrayList<>();
if (s == null || s.length() == 0 || wordDict == null || wordDict.size() == 0) {
return res;
}

int len = s.length();
List<String>[] dp = new ArrayList[len + 1];
dp[0] = new ArrayList<String>();

for (int i = 1; i <= len; i++) {
for (int j = i - 1; j >= 0; j--) {
String word = s.substring(j, i);
if (!wordDict.contains(word) || dp[j] == null) {
continue;
}
if (dp[i] == null) {
dp[i] = new ArrayList<String>();
}
dp[i].add(word);
}
}

if (dp[len] == null) {
return res;
}

String sol = "";
helper(res, sol, dp, len);
return res;
}

private void helper(List<String> res, String sol, List<String>[] dp, int end) {
if (end < 0) {
return ;
}
if (end == 0) {
String solution = new String(sol);
res.add(solution.trim());
return ;
}

for (String word : dp[end]) {
helper(res, word + " " + sol, dp, end - word.length());
}
}
}