140. Word Break II 🍺

README.md

@@ -129,6 +129,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 137 |[Single Number II](https://leetcode.com/problems/single-number-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/137.%20Single%20Number%20II.md)|Medium|  |
 | 138 |[Copy List with Random Pointer](https://leetcode.com/problems/copy-list-with-random-pointer/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/138.%20Copy%20List%20with%20Random%20Pointer.md)|Medium|  |
 | 139 |[Word Break](https://leetcode.com/problems/word-break/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/139.%20Word%20Break.md)|Medium|  |
+| 140 |[Word Break II](https://leetcode.com/problems/word-break-ii/)|[C++](solutions/140.%20Word%20Break%20II.md)|Hard|  |
 | 141 |[Linked List Cycle](https://leetcode.com/problems/linked-list-cycle)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/141.%20Linked%20List%20Cycle.md)|Easy|  |
 | 142 |[Linked List Cycle II](https://leetcode.com/problems/linked-list-cycle-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/142.%20Linked%20List%20Cycle%20II.md)|Medium|  |
 | 143 |[Reorder List](https://leetcode.com/problems/reorder-list/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/143.%20Reorder%20List.md)|Medium|  |

solutions/140. Word Break II.md

@@ -0,0 +1,41 @@
+# [140. Word Break II](https://leetcode.com/problems/word-break-ii/)
+
+# 思路
+
+这题是139. Word Break的升级版，139题只让我们判断给定的字符串s是否可以拆分成字典中的词，而这题让我们求出所有可以拆分成的情况。139题我们用的是动态规划或者带记忆的递归，此题其实也是类似的。我们用一个cache来记录已经求过的结果，即`cache[s]`就等于s被拆分的所有情况，那么我们就可根据`cache[s去掉某个前缀后的子串]`的结果求得`cache[s]`，伪代码如下：
+
+```
+cache[s] = {}
+for word in wordDict:
+    if word是s的前缀:
+        sub_res = s去掉前缀word后的拆分结果(即递归调用)
+        for ss in subres:
+            cache[s].push(word + " " + ss)
+```
+
+# C++
+``` C++
+class Solution {
+private:
+    unordered_map<string, vector<string>>cache;
+    vector<string> helper(string s, const vector<string>& wordDict){
+        if(s.empty()) return {""}; // 注意不是空数组, 因为如果为空就代表了无法拆分
+        if(cache.count(s)) return cache[s];
+        
+        vector<string>res;
+        for(string word: wordDict){
+            if(s.substr(0, word.size()) != word) continue;
+            vector<string>subres = helper(s.substr(word.size()), wordDict);
+            for(string &ss: subres)
+                res.push_back(word + (ss.empty() ? "" : " ") + ss);
+        }
+        // 如果不能被拆分res将为空数组
+        cache[s] = res;
+        return res;
+    }
+public:
+    vector<string> wordBreak(string s, vector<string>& wordDict) {
+        return helper(s, wordDict);
+    }
+};
+```