LeetCode140. Word Break II

题目：

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.

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"].

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

题目分析：

这道题和LeetCode139.Wordbreak差不多，但是更难一点，在动态规划的前提下，还得加入回溯(递归)。

而且我表达能力也不是特别好，这道题理解起来也不是特别难。但是看代码更简单直观。

代码：

class Solution {public:void breakword(string &s, vector<string>& wordDict, string tmp, int id, vector<bool>& dp,vector<string>& res) {for (int len = 1; len + id <= s.size(); len++) {if (len + id < s.size()) {if (dp[id] && find(wordDict.begin(), wordDict.end(), s.substr(id, len)) != wordDict.end()) {breakword(s, wordDict, tmp + s.substr(id, len) + " ", id + len, dp,res);}}else {if (dp[id] && find(wordDict.begin(), wordDict.end(), s.substr(id, len)) != wordDict.end())res.push_back(tmp + s.substr(id, len));}}}vector<string> wordBreak(string s, vector<string>& wordDict) {vector<bool> dp(s.size() + 1);dp[0] = 1;for (int i = 1; i <= s.size(); i++) {for (int j = i - 1; j >= 0; j--) {if (dp[j] && find(wordDict.begin(), wordDict.end(), s.substr(j, i - j)) != wordDict.end()) {dp[i] = 1;}}}vector<string> res;if (dp[s.size()]) {breakword(s, wordDict, "", 0, dp, res);return res;}else {return res;}}};