394. Decode String (Medium)

https://leetcode.com/problems/decode-string/

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".

Solutions

class Solution {

    public String decodeString(String s) {
        if (s == null || s.length() == 0) {
            return s;
        }

        Stack<Character> stack = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);

            if (c != ']') {
                stack.push(c);

                continue;
            }

            // c == ']'

            String pattern = "";

            // pop out elements in []
            while (stack.peek() != '[') {
                pattern = stack.pop() + pattern;
            }

            // drop '['
            stack.pop();

            String multiple = "";

            // retrieve the repeating times
            while (!stack.isEmpty() && stack.peek() >= '0' && stack.peek() <= '9') {
                multiple = stack.pop() + multiple;
            }

            // push in the repeated result string
            String tmp = repeat(pattern, multiple);
            for (char x : tmp.toCharArray()) {
                stack.push(x);
            }
        }

        String ans = "";
        while (!stack.isEmpty()) {
            ans = stack.pop() + ans;
        }

        return ans;
    }

    private String repeat(String pattern, String n) {
        String ans = "";

        if (n == null || n.length() == 0) {
            return ans;
        }

        for (int i = 0; i < Integer.valueOf(n); i++) {
            ans += pattern;
        }

        return ans;
    }
}

Incorrect Solutions

class Solution {

    // Incorrect solution, the bracket can be nested.

    public String decodeString(String s) {
        if (s == null || s.length() == 0) {
            return s;
        }

        Stack<Character> stack = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);

            if (c != ']') {
                stack.push(c);

                continue;
            }

            // c == ']'

            String pattern = "";

            // pop out elements in []
            while (stack.peek() != '[') {
                pattern = stack.pop() + pattern;
            }

            // drop '['
            stack.pop();

            String multiple = "";

            // retrieve the repeating times
            while (!stack.isEmpty() && stack.peek() >= '0' && stack.peek() <= '9') {
                multiple = stack.pop() + multiple;
            }

            // push in the repeated result string
            String tmp = repeat(pattern, multiple);
            for (char x : tmp.toCharArray()) {
                stack.push(x);
            }
        }

        String ans = "";
        while (!stack.isEmpty()) {
            ans = stack.pop() + ans;
        }

        return ans;
    }

    private String repeat(String pattern, String n) {
        String ans = "";

        if (n == null || n.length() == 0) {
            return ans;
        }

        for (int i = 0; i < Integer.valueOf(n); i++) {
            ans += pattern;
        }

        return ans;
    }
}

References

Copyright © iovi.com 2017 all right reserved,powered by GitbookLast Modification: 2019-12-03 11:01:17

results matching ""

    No results matching ""