Decode String

最新推荐文章于 2020-05-28 23:09:19 发布

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本文介绍了一种解决特定字符串解码问题的算法。该算法通过递归地寻找并处理字符串中的括号配对来实现字符串的解码。文章提供了一个详细的Java实现示例，包括如何定位左右括号、解析重复次数及递归解码内部字符串。

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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;
    }
}