Add solution 127

Author: yanglbme

README.md

@@ -36,7 +36,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 235 | [Lowest Common Ancestor of a Binary Search Tree](https://github.com/doocs/leetcode/tree/master/solution/235.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree) | `Tree` |
 | 237 | [Delete Node in a Linked List](https://github.com/doocs/leetcode/tree/master/solution/237.Delete%20Node%20in%20a%20Linked%20List) | `Linked List` |
 | 344 | [Reverse String](https://github.com/doocs/leetcode/tree/master/solution/344.Reverse%20String) | `Two Pointers`, `String` |
-| 703 | [Kth Largest Element in a Stream](%20ttps://github.com/doocs/leetcode/tree/master/solution/703.Kth%20Largest%20Element%20in%20a%20Stream) | `Heap` |
+| 703 | [Kth Largest Element in a Stream](https://github.com/doocs/leetcode/tree/master/solution/703.Kth%20Largest%20Element%20in%20a%20Stream) | `Heap` |
 | 876 | [Middle of the Linked List](https://github.com/doocs/leetcode/tree/master/solution/876.Middle%20of%20the%20Linked%20List) | `Linked List` |
 
 
@@ -58,6 +58,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 086 | [Partition List](https://github.com/doocs/leetcode/tree/master/solution/086.Partition%20List) | `Linked List`, `Two Pointers` |
 | 092 | [Reverse Linked List II](https://github.com/doocs/leetcode/tree/master/solution/092.Reverse%20Linked%20List%20II) | `Linked List` |
 | 094 | [Binary Tree Inorder Traversal](https://github.com/doocs/leetcode/tree/master/solution/094.Binary%20Tree%20Inorder%20Traversal) | `Hash Table`, `Stack`, `Tree` |
+| 127 | [Word Ladder](https://github.com/doocs/leetcode/tree/master/solution/127.Word%20Ladder) | `Breadth-first Search` |
 | 144 | [Binary Tree Preorder Traversal](https://github.com/doocs/leetcode/tree/master/solution/144.Binary%20Tree%20Preorder%20Traversal) | `Stack`, `Tree` |
 | 150 | [Evaluate Reverse Polish Notation](https://github.com/doocs/leetcode/tree/master/solution/150.Evaluate%20Reverse%20Polish%20Notation) | `Stack` |
 | 153 | [Find Minimum in Rotated Sorted Array](https://github.com/doocs/leetcode/tree/master/solution/153.Find%20Minimum%20in%20Rotated%20Sorted%20Array) | `Array`, `Binary Search` |
@@ -81,7 +82,7 @@ I'm looking for long-term contributors/partners to this repo! Send me PRs if you
 ## Contributors
 
 <!-- ALL-CONTRIBUTORS-LIST:START - Do not remove or modify this section -->
-| <center> [<img src="https://avatars3.githubusercontent.com/u/21008209?v=4" width="100px;"/><br /><sub><b>doocs</b></sub>](https://github.com/doocs)<br />[💻](https://github.com/doocs/leetcode/commits?author=doocs "Code") </center> | <center> [<img src="https://avatars3.githubusercontent.com/u/23625436?v=4" width="100px;"/><br /><sub><b>chakyam</b></sub>](https://github.com/chakyam)<br />[💻](https://github.com/doocs/leetcode/commits?author=chakyam "Code") </center> | 
+| <center> [<img src="https://avatars3.githubusercontent.com/u/21008209?v=4" width="100px;"/><br /><sub><b>yanglbme</b></sub>](https://github.com/yanglbme)<br />[💻](https://github.com/doocs/leetcode/commits?author=yanglbme "Code") </center> | <center> [<img src="https://avatars3.githubusercontent.com/u/23625436?v=4" width="100px;"/><br /><sub><b>chakyam</b></sub>](https://github.com/chakyam)<br />[💻](https://github.com/doocs/leetcode/commits?author=chakyam "Code") </center> | 
 |---|---|
 
 <!-- ALL-CONTRIBUTORS-LIST:END -->

solution/127.Word Ladder/README.md

@@ -0,0 +1,98 @@
+## 单词接龙
+### 题目描述
+
+给定两个单词（beginWord 和 endWord）和一个字典，找到从 beginWord 到 endWord 的最短转换序列的长度。转换需遵循如下规则：
+
+- 每次转换只能改变一个字母。
+- 转换过程中的中间单词必须是字典中的单词。
+
+说明:
+
+- 如果不存在这样的转换序列，返回 0。
+- 所有单词具有相同的长度。
+- 所有单词只由小写字母组成。
+- 字典中不存在重复的单词。
+- 你可以假设 beginWord 和 endWord 是非空的，且二者不相同。
+
+示例 1:
+```
+输入:
+beginWord = "hit",
+endWord = "cog",
+wordList = ["hot","dot","dog","lot","log","cog"]
+
+输出: 5
+
+解释: 一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog",
+     返回它的长度 5。
+```
+
+示例 2:
+```
+输入:
+beginWord = "hit"
+endWord = "cog"
+wordList = ["hot","dot","dog","lot","log"]
+
+输出: 0
+
+解释: endWord "cog" 不在字典中，所以无法进行转换。
+```
+
+### 解法
+利用广度优先搜索，`level` 表示层数，`curNum` 表示当前层的单词数，`nextNum` 表示下一层的单词数。
+队首元素出队，对于该字符串，替换其中每一个字符为[a...z] 中的任一字符。判断新替换后的字符串是否在字典中。若是，若该字符串与 endWord 相等，直接返回 level + 1；若不等，该字符串入队，nextNum + 1，并将 该字符串从 wordSet 移除。
+
+注意，每次只能替换一个字符，因此，在下一次替换前，需恢复为上一次替换前的状态。
+
+```java
+class Solution {
+    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
+        Queue<String> queue = new LinkedList<>();
+        
+        // 需要转hashSet
+        Set<String> wordSet = new HashSet<>(wordList);
+        queue.offer(beginWord);
+        int level = 1;
+        int curNum = 1;
+        int nextNum = 0;
+        while (!queue.isEmpty()) {
+            String s = queue.poll();
+            --curNum;
+            char[] chars = s.toCharArray();
+            for (int i = 0; i < chars.length; ++i) {
+                char ch = chars[i];
+                for (char j = 'a'; j <= 'z'; ++j) {
+                    chars[i] = j;
+                    String tmp = new String(chars);
+                    // 字典中包含生成的中间字符串
+                    if (wordSet.contains(tmp)) {
+                        // 中间字符串与 endWord 相等
+                        if (endWord.equals(tmp)) {
+                            return level + 1;
+                        }
+
+                        // 中间字符串不是 endWord，则入队
+                        queue.offer(tmp);
+                        ++nextNum;
+
+                        // 确保之后不会再保存 tmp 字符串
+                        wordSet.remove(tmp);
+                    }
+                }
+                chars[i] = ch;
+            }
+
+            if (curNum == 0) {
+                curNum = nextNum;
+                nextNum = 0;
+                ++level;
+            }
+
+        }
+        
+        return 0;
+    }
+}
+
+```

solution/127.Word Ladder/Solution.java

@@ -0,0 +1,48 @@
+class Solution {
+    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
+        Queue<String> queue = new LinkedList<>();
+        
+        // 需要转hashSet
+        Set<String> wordSet = new HashSet<>(wordList);
+        queue.offer(beginWord);
+        int level = 1;
+        int curNum = 1;
+        int nextNum = 0;
+        while (!queue.isEmpty()) {
+            String s = queue.poll();
+            --curNum;
+            char[] chars = s.toCharArray();
+            for (int i = 0; i < chars.length; ++i) {
+                char ch = chars[i];
+                for (char j = 'a'; j <= 'z'; ++j) {
+                    chars[i] = j;
+                    String tmp = new String(chars);
+                    // 字典中包含生成的中间字符串
+                    if (wordSet.contains(tmp)) {
+                        // 中间字符串与 endWord 相等
+                        if (endWord.equals(tmp)) {
+                            return level + 1;
+                        }
+
+                        // 中间字符串不是 endWord，则入队
+                        queue.offer(tmp);
+                        ++nextNum;
+
+                        // 确保之后不会再保存 tmp 字符串
+                        wordSet.remove(tmp);
+                    }
+                }
+                chars[i] = ch;
+            }
+
+            if (curNum == 0) {
+                curNum = nextNum;
+                nextNum = 0;
+                ++level;
+            }
+
+        }
+        
+        return 0;
+    }
+}