feat: add 017

Author: Blankj

README.md

@@ -61,6 +61,7 @@
 |3|[Longest Substring Without Repeating Characters][003]|Hash Table, Two Pointers, String|
 |8|[String to Integer (atoi)][008]|Math, String|
 |15|[3Sum][015]|Array, Two Pointers|
+|17|[Letter Combinations of a Phone Number][017]|String, Backtracking|
 |19|[Remove Nth Node From End of List][019]|Linked List, Two Pointers|
 |554|[Brick Wall][554]|Hash Table|
 
@@ -117,6 +118,7 @@
 [003]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/003/README.md
 [008]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/008/README.md
 [015]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/015/README.md
+[017]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/017/README.md
 [019]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/019/README.md
 [554]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/554/README.md
 

note/017/README.md

@@ -0,0 +1,85 @@
+# [Letter Combinations of a Phone Number][title]
+
+## Description
+
+Given a digit string, return all possible letter combinations that the number could represent.
+
+A mapping of digit to letters (just like on the telephone buttons) is given below.
+
+![img](https://upload.wikimedia.org/wikipedia/commons/thumb/7/73/Telephone-keypad2.svg/200px-Telephone-keypad2.svg.png)
+
+```
+Input:Digit string "23"
+Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
+
+```
+
+**Note:**
+Although the above answer is in lexicographical order, your answer could be in any order you want.
+
+**Tags:** String, Backtracking
+
+
+## 思路0
+
+题意是给你按键，让你组合出所有不同结果，首先想到的肯定是回溯了，对每个按键的所有情况进行回溯，回溯的终点就是结果字符串长度和按键长度相同。
+
+``` java
+class Solution {
+    private static String[] map = new String[]{"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
+
+    public List<String> letterCombinations(String digits) {
+        if (digits.length() == 0) return Collections.emptyList();
+        List<String> list = new ArrayList<>();
+        helper(list, digits, "");
+        return list;
+    }
+
+    private void helper(List<String> list, String digits, String ans) {
+        if (ans.length() == digits.length()) {
+            list.add(ans);
+            return;
+        }
+        for (char c : map[digits.charAt(ans.length()) - '2'].toCharArray()) {
+            helper(list, digits, ans + c);
+        }
+    }
+}
+```
+
+## 思路1
+
+还有一种思路就是利用队列，根据上一次队列中的值，该值拼接当前可选值来不断迭代其结果，具体代码如下。
+
+``` java
+class Solution {
+    private static String[] map = new String[]{"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
+
+    public List<String> letterCombinations(String digits) {
+        if (digits.length() == 0) return Collections.emptyList();
+        LinkedList<String> list = new LinkedList<>();
+        list.add("");
+        char[] charArray = digits.toCharArray();
+        for (int i = 0; i < charArray.length; i++) {
+            char c = charArray[i];
+            while (list.getFirst().length() == i) {
+                String pop = list.removeFirst();
+                for (char v : map[c - '2'].toCharArray()) {
+                    list.addLast(pop + v);
+                }
+            }
+        }
+        return list;
+    }
+}
+```
+
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl]
+
+
+
+[title]: https://leetcode.com/problems/letter-combinations-of-a-phone-number
+[ajl]: https://github.com/Blankj/awesome-java-leetcode

src/com/blankj/medium/_015/Solution.java

@@ -8,7 +8,7 @@
  * <pre>
  *     author: Blankj
  *     blog  : http://blankj.com
- *     time  : 2017/04/23
+ *     time  : 2017/10/14
  *     desc  :
  * </pre>
  */

src/com/blankj/medium/_017/Solution.java

@@ -0,0 +1,59 @@
+package com.blankj.medium._017;
+
+import java.util.Collections;
+import java.util.LinkedList;
+import java.util.List;
+
+/**
+ * <pre>
+ *     author: Blankj
+ *     blog  : http://blankj.com
+ *     time  : 2017/10/15
+ *     desc  :
+ * </pre>
+ */
+public class Solution {
+//    private static String[] map = new String[]{"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
+//
+//    public List<String> letterCombinations(String digits) {
+//        if (digits.length() == 0) return Collections.emptyList();
+//        List<String> list = new ArrayList<>();
+//        helper(list, digits, "");
+//        return list;
+//    }
+//
+//    private void helper(List<String> list, String digits, String ans) {
+//        if (ans.length() == digits.length()) {
+//            list.add(ans);
+//            return;
+//        }
+//        for (char c : map[digits.charAt(ans.length()) - '2'].toCharArray()) {
+//            helper(list, digits, ans + c);
+//        }
+//    }
+
+    private static String[] map = new String[]{"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
+
+    public List<String> letterCombinations(String digits) {
+        if (digits.length() == 0) return Collections.emptyList();
+        LinkedList<String> list = new LinkedList<>();
+        list.add("");
+        char[] charArray = digits.toCharArray();
+        for (int i = 0; i < charArray.length; i++) {
+            char c = charArray[i];
+
+            while (list.getFirst().length() == i) {
+                String pop = list.removeFirst();
+                for (char v : map[c - '2'].toCharArray()) {
+                    list.addLast(pop + v);
+                }
+            }
+        }
+        return list;
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println(solution.letterCombinations("23"));
+    }
+}