复原IP地址

Author: Folgerjun

src/com/leetcode_cn/medium/RestoreIPAddresses.java

@@ -0,0 +1,139 @@
+package com.leetcode_cn.medium;
+
+import java.util.ArrayList;
+import java.util.LinkedList;
+import java.util.List;
+
+/*****************复原IP地址************/
+/**
+ * 给定一个只包含数字的字符串，复原它并返回所有可能的 IP 地址格式。
+ * 
+ * 示例:
+ * 
+ * 输入: "25525511135"
+ * 
+ * 输出: ["255.255.11.135", "255.255.111.35"]
+ * 
+ * @author ffj
+ *
+ */
+public class RestoreIPAddresses {
+
+	// 数字长度
+	int n;
+	// 数字
+	String s;
+	// 存储三段数字
+	LinkedList<String> segments = new LinkedList<String>();
+	// 存储结果
+	ArrayList<String> output = new ArrayList<String>();
+
+	public List<String> restoreIpAddresses(String s) {
+		n = s.length();
+		this.s = s;
+		backtrack(-1, 3);
+		return output;
+	}
+
+	/**
+	 * 校验数字是否合理
+	 * 
+	 * @param segment
+	 * @return
+	 */
+	public boolean valid(String segment) {
+		/*
+		 * Check if the current segment is valid : 1. less or equal to 255 2. the first
+		 * character could be '0' only if the segment is equal to '0'
+		 */
+		int m = segment.length();
+		if (m > 3)
+			return false;
+		return (segment.charAt(0) != '0') ? (Integer.valueOf(segment) <= 255) : (m == 1);
+	}
+
+	/**
+	 * 分为三段
+	 * 
+	 * @param curr_pos
+	 */
+	public void update_output(int curr_pos) {
+		/*
+		 * Append the current list of segments to the list of solutions
+		 */
+		String segment = s.substring(curr_pos + 1, n);
+		if (valid(segment)) {
+			segments.add(segment);
+			// 三段拼接
+			output.add(String.join(".", segments));
+			// 回溯
+			segments.removeLast();
+		}
+	}
+
+	/**
+	 * 回溯法 一段一段截
+	 * 
+	 * @param prev_pos
+	 *            截取起始点
+	 * @param dots
+	 *            剩余段数
+	 */
+	public void backtrack(int prev_pos, int dots) {
+		/*
+		 * prev_pos : the position of the previously placed dot dots : number of dots to
+		 * place
+		 */
+		// The current dot curr_pos could be placed
+		// in a range from prev_pos + 1 to prev_pos + 4.
+		// The dot couldn't be placed
+		// after the last character in the string.
+		int max_pos = Math.min(n - 1, prev_pos + 4);
+		for (int curr_pos = prev_pos + 1; curr_pos < max_pos; curr_pos++) {
+			String segment = s.substring(prev_pos + 1, curr_pos + 1);
+			if (valid(segment)) {
+				segments.add(segment); // place dot
+				if (dots - 1 == 0) // if all 3 dots are placed
+					update_output(curr_pos); // add the solution to output
+				else
+					backtrack(curr_pos, dots - 1); // continue to place dots
+				segments.removeLast(); // remove the last placed dot
+			}
+		}
+	}
+
+	/**
+	 * 讨论中解法
+	 * 
+	 * 3-loop divides the string s into 4 substring: s1, s2, s3, s4. Check if each
+	 * substring is valid. In isValid, strings whose length greater than 3 or equals
+	 * to 0 is not valid; or if the string's length is longer than 1 and the first
+	 * letter is '0' then it's invalid; or the string whose integer representation
+	 * greater than 255 is invalid.
+	 * 
+	 * @param s
+	 * @return
+	 */
+	public List<String> restoreIpAddresses1(String s) {
+		List<String> res = new ArrayList<String>();
+		int len = s.length();
+		for (int i = 1; i < 4 && i < len - 2; i++) {
+			for (int j = i + 1; j < i + 4 && j < len - 1; j++) {
+				for (int k = j + 1; k < j + 4 && k < len; k++) {
+					String s1 = s.substring(0, i), s2 = s.substring(i, j), s3 = s.substring(j, k),
+							s4 = s.substring(k, len);
+					if (isValid(s1) && isValid(s2) && isValid(s3) && isValid(s4)) {
+						res.add(s1 + "." + s2 + "." + s3 + "." + s4);
+					}
+				}
+			}
+		}
+		return res;
+	}
+
+	public boolean isValid(String s) {
+		if (s.length() > 3 || s.length() == 0 || (s.charAt(0) == '0' && s.length() > 1) || Integer.parseInt(s) > 255)
+			return false;
+		return true;
+	}
+}