力扣算法——140WordBreakII【H】

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. Return all such possible sentences.

Note:

The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.

Example 1:

Input:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
  "cats and dog",
  "cat sand dog"
]

Example 2:

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]

Solution:
　　方法一，使用递归：【通不过牛客的例子】

 1 class Solution {
 2 public:
 3     unordered_map<string, vector<string>>map;
 4     vector<string> wordBreak(string s, unordered_set<string> &dict) {
 5         if (map.find(s) != map.end())return map[s];
 6         if (s.empty())return { "" };
 7         vector<string>res;
 8         for (auto word : dict)
 9         {
10             if (s.substr(0, word.size()) != word)continue;
11             vector<string>rem = wordBreak(s.substr(word.size()), dict);
12             for (auto str : rem)
13                 res.push_back(word + (str.empty() ? "" : " ") + str);
14         }
15         return map[s] = res;
16     }
17 };




　　方法二：使用动态规划

 1 //使用动态规划
 2 
 3 class Solution {
 4 public:
 5     vector<string> wordBreak(string s, unordered_set<string> &dict) {
 6         int len = s.length();
 7         dp = new vector<bool>[len];
 8         for (int pos = 0; pos < len; pos++) {
 9             for (int i = 1; i < len - pos + 1; i++) {
10                 if (dict.find(s.substr(pos, i)) != dict.end())
11                     dp[pos].push_back(true);
12                 else
13                     dp[pos].push_back(false);
14             }
15         }
16         dfs(s, len - 1);
17         return res;
18     }
19     void dfs(string s, int n) {
20         if (n >= 0) {
21             for (int i = n; i >= 0; i--) {
22                 if (dp[i][n - i]) {
23                     mid.push_back(s.substr(i, n - i + 1));
24                     dfs(s, i - 1);
25                     mid.pop_back();
26                 }
27             }
28         }
29         else {
30             string r;
31             for (int j = mid.size() - 1; j >= 0; j--) {
32                 r += mid[j];
33                 if (j > 0)
34                     r += " ";
35             }
36             res.push_back(r);
37         }
38     }
39     vector<bool> *dp;
40     vector<string> res;
41     vector<string> mid;
42 };