Word Break II 解答

Question

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

Solution

这题的基本思路是用recursion，每次都只研究第一刀切在哪儿，然后对剩下的substring做递归。但是我们发现其实可以借助DP的思想，存下来部分中间结果值。这种思想也叫 Memorized Recursion。

Memorized recursion并不是一个很好的方法，但是当遇到time limit exceed时可以想到它来拿空间换时间

class Solution:
    # @param s, a string
    # @param dict, a set of string
    # @return a list of strings
    def wordBreak(self, s, dict):
        self.dict = dict
        # cache stores tmp results: key: substring value: list of results
        self.cache = {}
        return self.break_helper(s)

    def break_helper(self, s):
        combs = []
        if s in self.cache:
            return self.cache[s]
        if len(s) == 0:
            return []

        for i in range(len(s)):
            if s[:i+1] in self.dict:
                if i == len(s) - 1:
                    combs.append(s[:i+1])
                else:
                    sub_combs = self.break_helper(s[i+1:])
                    for sub_comb in sub_combs:
                        combs.append(s[:i+1] + ' ' + sub_comb)

        self.cache[s] = combs
        return combs

第二种思路：先用DP，再用DFS

dp[i] 为list<String>表示以i结尾的在word dict中的词，并且满足dp[i - wordLen]不是null。

DFS从dp[len]开始，遍历可能的词。

public class Solution {
    public List<String> wordBreak(String s, Set<String> wordDict) {
        List<String> result = new ArrayList<>();
        if (wordDict == null || s == null) {
            return result;
        }
        int len = s.length();
        ArrayList<String>[] dp = new ArrayList[len + 1];
        dp[0] = new ArrayList<String>();
        for (int i = 0; i < len; i++) {
            if (dp[i] != null) {
                for (int j = i + 1; j <= len; j++) {
                    String sub = s.substring(i, j);
                    if (wordDict.contains(sub)) {
                        if (dp[j] == null) {
                            dp[j] = new ArrayList<String>();
                        }
                        dp[j].add(sub);
                    }
                }
            }
        }
        if (dp[len] == null) {
            return result;
        }
        // dfs
        dfs(dp, result, "", len);
        return result;
    }
    
    private void dfs(ArrayList<String>[] dp, List<String> result, String tail, int curPos) {
        if (curPos == 0) {
            result.add(tail.trim());
            return;
        }
        for (String str : dp[curPos]) {
            String curStr = str + " " + tail;
            dfs(dp, result, curStr, curPos - str.length());
        }
    }
}