Decode String

描述

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"

分析

这题的麻烦之处在于如何处理嵌套结构，例如 k[string k[string]]，需要用栈或者递归。

代码

Python

// Decode String
// DFS
// Time Complexity:O(maxK*n), Space Complexity: O(n)
class Solution {
    private int i = 0;

    public String decodeString(String s) {
        StringBuilder result = new StringBuilder();

        while (i < s.length() && s.charAt(i) != ']') {
            if (!Character.isDigit(s.charAt(i))) {
                result.append(s.charAt(i++));
            } else {
                // parse the number
                int n = 0;
                while (i < s.length() && Character.isDigit(s.charAt(i)))
                    n = n * 10 + s.charAt(i++) - '0';

                i++; // ignore the opening bracket '['
                String decodedString = decodeString(s);
                i++; // ignore the closing bracket ']'

                // repeat n times
                while (n-- > 0) result.append(decodedString);
            }
        }
        return result.toString();
    }
}

Java

// TODO

C++

# Decode String
# DFS
# Time Complexity:O(maxK*n), Space Complexity: O(n)
class Solution:
    def __init__(self):
        self.i = 0

    def decodeString(self, s: str) -> str:
        result = []

        while self.i < len(s) and s[self.i] != ']':
            if not s[self.i].isdigit():
                result.append(s[self.i])
                self.i += 1
            else:
                # parse the number
                n = 0
                while self.i < len(s) and s[self.i].isdigit():
                    n = n * 10 + int(s[self.i])
                    self.i += 1

                self.i += 1  # ignore the opening bracket '['
                decoded_string = self.decodeString(s)
                self.i += 1  # ignore the closing bracket ']'

                # repeat n times
                result.append(decoded_string * n)

        return ''.join(result)