20180218

code/lc105.java

@@ -37,5 +37,4 @@ public class lc105 {
         return tn;
     }
 
-
 }

code/lc127.java

+package code;
+/*
+ * 127. Word Ladder
+ * 题意：给定开始字符串，结束字符串，和一个字符串数组，每次替换字符串中的一个字符，问最少几个步骤变为终止字符串
+ * 难度：Medium
+ * 分类：Breadth-first Search
+ * 思路：bfs, 利用双向bfs可以加快搜索https://leetcode.com/problems/word-ladder/discuss/40711/Two-end-BFS-in-Java-31ms.
+ * Tips：拓扑排序，很经典的BFS，好好看看
+ */
+import java.util.ArrayDeque;
+import java.util.List;
+import java.util.Queue;
+
+public class lc127 {
+    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
+        if(!wordList.contains(endWord)) return 0;
+        Queue<String> qu = new ArrayDeque();    //用一个Queue和int size类似树的层次遍历，和两个hashset效果一样
+        qu.add(beginWord);
+        int level = 2;
+        while(!qu.isEmpty()){
+            int size = qu.size();
+            for (int i = 0; i < size ; i++) {
+                char[] curr_str = qu.remove().toCharArray();
+                System.out.println(String.valueOf(curr_str));
+                for (int j = 0; j < curr_str.length ; j++) {
+                    char ch = curr_str[j];
+                    for (char k = 'a'; k <='z' ; k++) { //如果每次比较两个字符串是否差一位，时间复杂度太大，所以直接替换一个字符
+                        curr_str[j] = k;
+                        if(String.valueOf(curr_str).equals(endWord)) return level;
+                        if(wordList.contains(String.valueOf(curr_str))){
+                            wordList.remove(String.valueOf(curr_str));
+                            qu.add(String.valueOf(curr_str));
+                        }
+                    }
+                    curr_str[j] = ch;
+                }
+            }
+            level++;
+        }
+        return 0;
+    }
+}

readme.md

@@ -86,6 +86,7 @@ Language: Java
 | 122 [Java](./code/lc122.java)
 | 124 [Java](./code/lc124.java)
 | 125 [Java](./code/lc125.java)
+| 127 [Java](./code/lc127.java)
 | 128 [Java](./code/lc128.java)
 | 136 [Java](./code/lc136.java)
 | 139 [Java](./code/lc139.java)