0394. Decode String (M)

Decode String (M)

题目

Given an encoded string, return its 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].

Example 1:

Input: s = "3[a]2[bc]"
Output: "aaabcbc"

Example 2:

Input: s = "3[a2[c]]"
Output: "accaccacc"

Example 3:

Input: s = "2[abc]3[cd]ef"
Output: "abcabccdcdcdef"

Example 4:

Input: s = "abc3[cd]xyz"
Output: "abccdcdcdxyz"

Constraints:

• 1 <= s.length <= 30
• s consists of lowercase English letters, digits, and square brackets '[]'.
• s is guaranteed to be a valid input.
• All the integers in s are in the range [1, 300].

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题意

按要求展开给定的字符串。

思路

递归：从左到右遍历字符串，如果是字母则直接加入的结果中；如果是数字，则先计算数字，再递归展开'[ ]'中的字符串，最后拼接起来。

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代码实现

Java

class Solution {
    public String decodeString(String s) {
        String ans = "";
        int i = 0;

        while (i < s.length()) {
            char c = s.charAt(i);
            if (c < '0' || c > '9') {
                ans += c;
                i++;
            } else {
              	// 计算数字
                int count = 0;
                while (s.charAt(i) != '[') {
                    count = count * 10 + s.charAt(i++) - '0';
                }
                i++;
                int start = i;
              
                // 找到匹配的']'
                int leftCount = 0;
                while (i < s.length()) {
                    if (s.charAt(i) == '[') {
                        leftCount++;
                    } else if (s.charAt(i) == ']' && leftCount > 0) {
                        leftCount--;
                    } else if (s.charAt(i) == ']' && leftCount == 0) {
                        break;
                    }
                    i++;
                }

                String tmp = decodeString(s.substring(start, i));
                ans += tmp.repeat(count);
                
                i++;
            }
        }

        return ans;
    }
}