139. Word Break

使用动态规划的方法。public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
       
        boolean[] f = new boolean[s.length() + 1];
       
        f[0] = true;
       
       
        /* First DP
        for(int i = 1; i <= s.length(); i++){
            for(String str: dict){
                if(str.length() <= i){
                    if(f[i - str.length()]){
                        if(s.substring(i-str.length(), i).equals(str)){
                            f[i] = true;
                            break;
                        }
                    }
                }
            }
        }*/
       
        //Second DP
        for(int i=1; i <= s.length(); i++){
            for(int j=0; j < i; j++){
                if(f[j] && dict.contains(s.substring(j, i))){
                    f[i] = true;
                    break;
                }
            }
        }
       
        return f[s.length()];
    }
}

使用DFS的方法，注意使用了HashSet对扫描过的失败节点进行了标记。public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        // DFS
        Set<Integer> set = new HashSet<Integer>();
        return dfs(s, 0, dict, set);
    }
   
    private boolean dfs(String s, int index, Set<String> dict, Set<Integer> set){
        // base case
        if(index == s.length()) return true;
        // check memory
        if(set.contains(index)) return false;
        // recursion
        for(int i = index+1;i <= s.length();i++){
            String t = s.substring(index, i);
            if(dict.contains(t))
                if(dfs(s, i, dict, set))
                    return true;
                else
                    set.add(i);
        }
        set.add(index);
        return false;
    }
}