139. Word Break

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.

For example, given

s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

一刷
题解：
最简单的方法：dfs

class Solution {
    public boolean wordBreak(String s, List wordDict) {
        Set set = new HashSet<>(wordDict);
        
        return dfs(s, set, 0);
    }
    
    public boolean dfs(String s, Set set, int pos){
        if(pos == s.length()) return true;
        for(int i=pos+1; i<=s.length(); i++){
            if(set.contains(s.substring(pos, i))){
                //System.out.println(s.substring(pos, i));
                if(dfs(s, set, i)) return true;
            }
        }
        return false;
    }
}

但是在

"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]

会超时， O(n!)。考虑用dp来做：

O(n^2)

1. 可以直接由list转为set, 在构造set时传入list为构造函数参数
2. dynamic programming,
   dp[i] = dp[j] && set.contains(s.subString(i, j))
3. 要注意设置dp[0]=true, 对于DP来说，设置初始值都是十分重要的。

public class Solution {
    public boolean wordBreak(String s, List wordDict) {
        Set dict = new HashSet(wordDict);
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;//very important!!!
        for(int i=1; i<=s.length(); i++){
            for(int j = 0; j

二刷
DP

public class Solution {
    public boolean wordBreak(String s, List wordDict) {
        Set dict = new HashSet<>(wordDict);
        boolean[] dp = new boolean[s.length()+1];
        dp[0] = true;
        for(int right = 1; right<=s.length(); right++){
            for(int left = right-1; left>=0; left--){
                if(dp[left] && dict.contains(s.substring(left, right))) {
                    dp[right] = true;//not include right
                    break;
                }
            }
        }
        return dp[s.length()];
    }
}