[leetcode]Word Break

Word Break

 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".

算法思路：

思路1.dfs。肯定会超时，但是这里捡了leetcode的漏，本题case比较强的都是在结尾出现了坑，这道题偷懒，从后往前找单词。再递归。居然过了。。。。

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        if(s.length() == 0) return true;
        for(String word : dict){
            if(s.endsWith(word)){
                if(!wordBreak(s.substring(0, s.length() - word.length()),dict)) continue;
                else return true;
            }
        }
        return false;
    }
}

思路2：既然递归超时，那就用DP吧，dp[i]表示前i长度字符串能否成功分解。

代码如下：

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        int length = s.length();
        if(length == 0) return true;
        boolean[] dp = new boolean[length + 1];
        dp[0] = true;
        for(int i = 0; i <= length; i++){
            if(!dp[i]) continue;
            for(String word:dict){
                if(s.substring(i).startsWith(word)){
                    dp[i + word.length()] = true;
                }
            }
        }
        return dp[length];
    }
}