0093. Restore IP Addresses

coderbean · coderbean · commit ef9215732620 · 2020-02-29T15:49:00.000+08:00
diff --git a/docs/_sidebar.md b/docs/_sidebar.md
@@ -66,4 +66,5 @@
 * [0090. Subsets II](markdown/0090.%20Subsets%20II)
 * [0091. Decode Ways](markdown/0091.%20Decode%20Ways)
 * [0092. Reverse Linked List II](markdown/0092.%20Reverse%20Linked%20List%20II)
+* [0093. Restore IP Addresses](markdown/0093.%20Restore%20IP%20Addresses)
 * [0146. LRU Cache](markdown/0146.%20LRU%20Cache)
diff --git a/docs/markdown/0093. Restore IP Addresses.md b/docs/markdown/0093. Restore IP Addresses.md
@@ -0,0 +1,57 @@
+### [93\. Restore IP Addresses](https://leetcode.com/problems/restore-ip-addresses/)
+
+Difficulty: **Medium**
+
+
+Given a string containing only digits, restore it by returning all possible valid IP address combinations.
+
+**Example:**
+
+```
+Input: "25525511135"
+Output: ["255.255.11.135", "255.255.111.35"]
+```
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+class Solution {
+    public List<String> restoreIpAddresses(String s) {
+        List<String> result = new ArrayList<>();
+        doRestore(result, "", s, 0);
+        return result;
+    }
+​
+    /**
+     * DFS 反解析 IP 地址
+     *
+     * @param result 解析结果
+     * @param path   已经走完的路径：譬如：192.169
+     * @param s      剩下的需要解析的字符串
+     * @param level  DFS 的第几层
+     */
+    private void doRestore(List<String> result, String path, String s, int level) {
+        if ((4 - level) * 3 < s.length()) {
+            return;
+        }
+        // 如果大于三层，说明已经完成了
+        if (s.isEmpty()) {
+            if (level == 4) {
+                // 因为 path 最前面有一个 "." 需要去除
+                result.add(path.substring(1));
+            }
+            return;
+        }
+        // 开始的那个字母
+        String startNum = s.substring(0, 1);
+        for (int i = 0; i < (startNum.equals("0") ? 1 : (Math.min(s.length(), 3))); i++) {
+            if (Integer.parseInt(s.substring(0, i + 1)) <= 255) {
+                doRestore(result, path + "." + s.substring(0, i + 1), s.substring(i + 1), level + 1);
+            }
+        }
+    }
+}
+```
diff --git a/src/main/java/leetcode/_93_/Main.java b/src/main/java/leetcode/_93_/Main.java
@@ -0,0 +1,13 @@
+package leetcode._93_;
+
+/**
+ * Created by zhangbo54 on 2019-03-04.
+ */
+public class Main {
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println( solution.restoreIpAddresses("25525511135"));
+        System.out.println( solution.restoreIpAddresses("1111"));
+    }
+}
+
diff --git a/src/main/java/leetcode/_93_/Solution.java b/src/main/java/leetcode/_93_/Solution.java
@@ -0,0 +1,41 @@
+package leetcode._93_;
+
+import java.util.ArrayList;
+import java.util.List;
+
+class Solution {
+    public List<String> restoreIpAddresses(String s) {
+        List<String> result = new ArrayList<>();
+        doRestore(result, "", s, 0);
+        return result;
+    }
+
+    /**
+     * DFS 反解析 IP 地址
+     *
+     * @param result 解析结果
+     * @param path   已经走完的路径：譬如：192.169
+     * @param s      剩下的需要解析的字符串
+     * @param level  DFS 的第几层
+     */
+    private void doRestore(List<String> result, String path, String s, int level) {
+        if ((4 - level) * 3 < s.length()) {
+            return;
+        }
+        // 如果大于三层，说明已经完成了
+        if (s.isEmpty()) {
+            if (level == 4) {
+                // 因为 path 最前面有一个 "." 需要去除
+                result.add(path.substring(1));
+            }
+            return;
+        }
+        // 开始的那个字母
+        String startNum = s.substring(0, 1);
+        for (int i = 0; i < (startNum.equals("0") ? 1 : (Math.min(s.length(), 3))); i++) {
+            if (Integer.parseInt(s.substring(0, i + 1)) <= 255) {
+                doRestore(result, path + "." + s.substring(0, i + 1), s.substring(i + 1), level + 1);
+            }
+        }
+    }
+}
diff --git a/src/main/java/leetcode/_93_/solution.md b/src/main/java/leetcode/_93_/solution.md
@@ -0,0 +1,57 @@
+### [93\. Restore IP Addresses](https://leetcode.com/problems/restore-ip-addresses/)
+
+Difficulty: **Medium**
+
+
+Given a string containing only digits, restore it by returning all possible valid IP address combinations.
+
+**Example:**
+
+```
+Input: "25525511135"
+Output: ["255.255.11.135", "255.255.111.35"]
+```
+
+
+#### Solution
+
+Language: **Java**
+
+```java
+class Solution {
+    public List<String> restoreIpAddresses(String s) {
+        List<String> result = new ArrayList<>();
+        doRestore(result, "", s, 0);
+        return result;
+    }
+​
+    /**
+     * DFS 反解析 IP 地址
+     *
+     * @param result 解析结果
+     * @param path   已经走完的路径：譬如：192.169
+     * @param s      剩下的需要解析的字符串
+     * @param level  DFS 的第几层
+     */
+    private void doRestore(List<String> result, String path, String s, int level) {
+        if ((4 - level) * 3 < s.length()) {
+            return;
+        }
+        // 如果大于三层，说明已经完成了
+        if (s.isEmpty()) {
+            if (level == 4) {
+                // 因为 path 最前面有一个 "." 需要去除
+                result.add(path.substring(1));
+            }
+            return;
+        }
+        // 开始的那个字母
+        String startNum = s.substring(0, 1);
+        for (int i = 0; i < (startNum.equals("0") ? 1 : (Math.min(s.length(), 3))); i++) {
+            if (Integer.parseInt(s.substring(0, i + 1)) <= 255) {
+                doRestore(result, path + "." + s.substring(0, i + 1), s.substring(i + 1), level + 1);
+            }
+        }
+    }
+}
+```
