leetcode 394. Decode String
题目:
Given an encoded string, return it's decoded string.
The encoding rule is: k[encoded_string]
, where the encoded_string inside the square brackets is being repeated exactly k times. Note thatk is guaranteed to be a positive integer.
You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.
Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won't be input like 3a
or2[4]
.
Examples:
s = "3[a]2[bc]", return "aaabcbc". s = "3[a2[c]]", return "accaccacc". s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
1 import java.util.Stack; 2 import java.util.regex.Pattern; 3 4 public class Solution { 5 public String decodeString(String s) { 6 Stack<String> sta = new Stack<String>(); 7 char[] c = s.toCharArray(); 8 StringBuffer sb = new StringBuffer(); 9 StringBuffer result = new StringBuffer(); 10 for(int i = 0; i<c.length; i++){ 11 if(c[i]==']'){ 12 while(!(sta.peek()).equals("[")) { 13 sb.insert(0, sta.pop()); 14 } 15 sta.pop(); 16 int num = Integer.parseInt(sta.pop()); 17 String sbtemp = sb.toString(); 18 for(int j=0;j<num-1;j++){ 19 sb.append(sbtemp); 20 } 21 if(sta.size()==0){ 22 result.append(sb); 23 sb.setLength(0); 24 } 25 else { 26 sta.push(sb.toString()); 27 sb.setLength(0); 28 } 29 } 30 else { 31 Pattern pattern = Pattern.compile("[0-9]*"); 32 while(pattern.matcher(""+c[i]).matches()){ 33 sb.append(c[i]); 34 i++; 35 } 36 if(sb.length()>0) 37 sta.push(sb.toString()); 38 sta.push("" + c[i]); 39 sb.setLength(0); 40 } 41 } 42 while(sta.size()>0){ 43 sb.insert(0,sta.pop()); 44 } 45 return (result.append(sb)).toString(); 46 } 47 }