leetcode 394. Decode String

最新推荐文章于 2024-06-19 18:22:32 发布

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最新推荐文章于 2024-06-19 18:22:32 发布

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本文介绍了一种字符串解码算法的实现方法，通过两种不同的代码实现展示了如何将编码字符串（如3[a]2[bc]）转换为其解码形式（如aaabcbc）。文章提供了递归方法和使用栈进行迭代的方法，并详细解释了每一步的处理逻辑。

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".

我觉得这种题目都不是考逻辑思维，而是考编码能力的。。。

public String decodeString(String s) {
	if(s.indexOf('[')==-1){
		return s;
	}
	StringBuilder sb=new StringBuilder();
	char[] cs=s.toCharArray();
	String integs="";
	int i=0;
	while(i<cs.length){
		if(cs[i]>='0'&&cs[i]<='9'){
			integs+=cs[i];
			i++;
		}
		else if(cs[i]=='['){
			int leftBound=1;
			i++;
			String innerString="";
			while(i<cs.length&&leftBound>0){
				if(cs[i]=='['){
					leftBound++;
				}
				else if(cs[i]==']'){
					leftBound--;
				}
				if(leftBound>0){
					innerString+=cs[i];
				}
				i++;
			}
			String repeatString=decodeString(innerString);
			int repeatTimes=Integer.parseInt(integs);
			for(int count=0;count<repeatTimes;count++){
				sb.append(repeatString);
			}
			integs="";
		}
		else{//9[cs]ab中的ab这样的情况
			sb.append(cs[i]);
			i++;
		}
	}
	return sb.toString();
}

大神没用递归，用的栈来迭代。

public class Solution {
    public String decodeString(String s) {
        String res = "";
        Stack<Integer> countStack = new Stack<>();
        Stack<String> resStack = new Stack<>();
        int idx = 0;
        while (idx < s.length()) {
            if (Character.isDigit(s.charAt(idx))) {
                int count = 0;
                while (Character.isDigit(s.charAt(idx))) {
                    count = 10 * count + (s.charAt(idx) - '0');
                    idx++;
                }
                countStack.push(count);
            }
            else if (s.charAt(idx) == '[') {
                resStack.push(res);
                res = "";//当前的res存储当前的repeatString
                idx++;
            }
            else if (s.charAt(idx) == ']') {
                StringBuilder temp = new StringBuilder (resStack.pop());
                int repeatTimes = countStack.pop();
                for (int i = 0; i < repeatTimes; i++) {
                    temp.append(res);
                }
                res = temp.toString();
                idx++;
            }
            else {
                res += s.charAt(idx++);
            }
        }
        return res;
    }
}