Word Break II 解答

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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());
}
}
}