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;
    }
}