From a3ba386ad6037ffafb80e2dac466b0b37146e304 Mon Sep 17 00:00:00 2001
From: nanhuaibeian <49868746+nanhuaibeian@users.noreply.github.com>
Date: Wed, 12 May 2021 21:59:18 +0800
Subject: [PATCH] =?UTF-8?q?Update=200017.=E7=94=B5=E8=AF=9D=E5=8F=B7?=
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---
 problems/0017.电话号码的字母组合.md | 38 ++++++++++++++++++++
 1 file changed, 38 insertions(+)

diff --git a/problems/0017.电话号码的字母组合.md b/problems/0017.电话号码的字母组合.md
index 6f51a181..76d5655d 100644
--- a/problems/0017.电话号码的字母组合.md
+++ b/problems/0017.电话号码的字母组合.md
@@ -240,7 +240,45 @@ public:
 
 
 Java：
+```Java
+class Solution {
 
+    //设置全局列表存储最后的结果
+    List<String> list = new ArrayList<>();
+
+    public List<String> letterCombinations(String digits) {
+        if (digits == null || digits.length() == 0) {
+            return list;
+        }
+        //初始对应所有的数字，为了直接对应2-9，新增了两个无效的字符串""
+        String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
+        //迭代处理
+        backTracking(digits, numString, 0);
+        return list;
+
+    }
+
+    //每次迭代获取一个字符串，所以会设计大量的字符串拼接，所以这里选择更为高效的 StringBuild
+    StringBuilder temp = new StringBuilder();
+
+    //比如digits如果为"23",num 为0，则str表示2对应的 abc
+    public void backTracking(String digits, String[] numString, int num) {
+        //遍历全部一次记录一次得到的字符串
+        if (num == digits.length()) {
+            list.add(temp.toString());
+            return;
+        }
+        //str 表示当前num对应的字符串
+        String str = numString[digits.charAt(num) - '0'];
+        for (int i = 0; i < str.length(); i++) {
+            temp.append(str.charAt(i));
+            backTracking(digits, numString, num + 1);
+            //剔除末尾的继续尝试
+            temp.deleteCharAt(temp.length() - 1);
+        }
+    }
+}
+```
 
 Python：