394. Decode String

最新推荐文章于 2024-07-30 15:25:02 发布

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本文介绍了一种用于解码特定格式字符串的算法。该算法能够处理形如 k[encoded_string] 的编码规则，其中 encoded_string 会被重复 k 次。通过使用栈辅助处理括号内的字符串，递归地进行解码。

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".

扫描串s，单个字母直接decode，遇到数字num把num之后的最外层括号对应的串去掉括号之后递归decode，将递归得到的结果重复num遍添加到当前串的decode当中。取带括号的子结构使用栈做帮助，栈的功能可以简化地的用计数值代替。

public class Solution {
    public static String decodeString(String s)
	{
		int len=s.length();
		StringBuilder sb=new StringBuilder();
		
		for(int i=0;i<len;)
		{
			char c=s.charAt(i);
			if(c>='0'&&c<='9')
			{
				int start=i;
				while(i<len&&s.charAt(i)>='0'&&s.charAt(i)<='9')
					i++;
				int times=Integer.parseInt(s.substring(start, i));
				int left=1;
				start=i;
				while(i<len&&left>0)
				{
					i++;
					if(s.charAt(i)=='[')
						left++;
					else if(s.charAt(i)==']')
						left--;
				}
				
				String ts=decodeString(s.substring(start+1, i));
				for(int k=0;k<times;k++)
					sb.append(ts);
					
				i++;
			}
			else {
				sb.append(c);
				i++;
			}
		}
		
		return sb.toString();
	}
}