README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0394.Decode%20String/README_EN.md	Stack Recursion String

394. Decode String

中文文档

Description

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; there are 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 will not be input like 3a or 2[4].

The test cases are generated so that the length of the output will never exceed 105.

 

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"

 

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].

Solutions

Solution 1

Python3

class Solution:
    def decodeString(self, s: str) -> str:
        s1, s2 = [], []
        num, res = 0, ''
        for c in s:
            if c.isdigit():
                num = num * 10 + int(c)
            elif c == '[':
                s1.append(num)
                s2.append(res)
                num, res = 0, ''
            elif c == ']':
                res = s2.pop() + res * s1.pop()
            else:
                res += c
        return res

Java

class Solution {
    public String decodeString(String s) {
        Deque<Integer> s1 = new ArrayDeque<>();
        Deque<String> s2 = new ArrayDeque<>();
        int num = 0;
        String res = "";
        for (char c : s.toCharArray()) {
            if ('0' <= c && c <= '9') {
                num = num * 10 + c - '0';
            } else if (c == '[') {
                s1.push(num);
                s2.push(res);
                num = 0;
                res = "";
            } else if (c == ']') {
                StringBuilder t = new StringBuilder();
                for (int i = 0, n = s1.pop(); i < n; ++i) {
                    t.append(res);
                }
                res = s2.pop() + t.toString();
            } else {
                res += String.valueOf(c);
            }
        }
        return res;
    }
}

TypeScript

function decodeString(s: string): string {
    let ans = '';
    let stack = [];
    let count = 0; // repeatCount
    for (let cur of s) {
        if (/[0-9]/.test(cur)) {
            count = count * 10 + Number(cur);
        } else if (/[a-z]/.test(cur)) {
            ans += cur;
        } else if ('[' == cur) {
            stack.push([ans, count]);
            // reset
            ans = '';
            count = 0;
        } else {
            // match ']'
            let [pre, count] = stack.pop();
            ans = pre + ans.repeat(count);
        }
    }
    return ans;
}