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".
分析 ： 用两个stack
public class Solution { public String decodeString(String s) { String res = ""; Stack<String> charStack = new Stack<>(); Stack<Integer> numStack = new Stack<>(); char sArray[] = s.toCharArray(); for(int i = 0 ; i < s.length(); i++){ if(sArray[i] > ‘0‘ && sArray[i] <= ‘9‘){ // if num = 10,111... while int j = i; int count = 0; while(sArray[j] >= ‘0‘ && sArray[j] <= ‘9‘){ //考虑大于1位数的情况 count = count * 10 + sArray[j] - ‘0‘; j++; } i = j - 1; numStack.push(count); } else if(sArray[i] == ‘[‘){ charStack.push(res); res = ""; } else if(sArray[i] == ‘]‘){ StringBuilder temp = new StringBuilder(charStack.pop()); int count = numStack.pop(); while(count > 0){ temp.append(res); count--; } res = temp.toString(); } else res += sArray[i]; } return res; } }
原文：http://www.cnblogs.com/joannacode/p/6014638.html