Word Break

Description:

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

问题描述：

判断给定的字符串能否被给定字典中的词分隔开来。。

Ex:

givens = “leetcode”,
dict = [“leet”, “code”].

Return true because “leetcode” can be segmented as “leet code”.

解法一：

思路：

动态规划，用dp[i]表示s[0…i]能够被分隔表示，用两个循环来遍历所有可能的情况，这里用的trick是s.substring()方法为前闭后开型.
所以在if逻辑中用来判断分隔字符串的衔接问题。
一旦匹配到字符串则用break退出内循环。。

Code:

public class Solution {    public boolean wordBreak(String s, List<String> wordDict) {        if(s == null && wordDict == null)             return true;        if(s == null || wordDict == null)            return false;        //dp[i] represents if s.substring(0, i) is wordbreakable.        boolean[] dp =  new boolean[s.length() + 1];        dp[0] = true;        for(int i = 1; i <= s.length(); i++){            for(int j = 0; j < i; j++){                if(dp[j] && wordDict.contains(s.substring(j,i))){                    dp[i] = true;                    break;                }            }        }        return dp[s.length()];    }}