Decode String

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 that k 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 or 2[4].

Examples:

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

这个题先找左括号匹配的右括号，将括号中间的当作子问题用递归求解即可。

public class Solution {
    public String decodeString(String s) {
        int n=s.length();
        int i=0;
        StringBuilder sb=new StringBuilder();
        while(i<s.length()){
            if(s.charAt(i)<='9'&&s.charAt(i)>='0'){
                int left=findLeftSquare(s,i+1);
                int right=findRightSquare(s,left+1);
                int num=Integer.parseInt(s.substring(i, left));
                for(int j=0;j<num;j++){
                    sb.append(decodeString(s.substring(left+1, right)));
                }
                i=right+1;
            }else{
                sb.append(s.charAt(i));
                i++;
            }
        }
        return sb.toString();
    }

    private int findRightSquare(String s, int from) {
        // TODO Auto-generated method stub
        int leftSquare=1;
        while(leftSquare!=0){
            if(s.charAt(from)=='['){
                leftSquare++;
            }
            if(s.charAt(from)==']'){
                leftSquare--;
            }
            from++;
        }
        return from-1;
    }

    private int findLeftSquare(String s, int from) {
        // TODO Auto-generated method stub
        while(s.charAt(from)!='['){
            from++;
        }
        return from;
    }
}