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  • [LC] 394. Decode String

    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].

    Examples:

    s = "3[a]2[bc]", return "aaabcbc".
    s = "3[a2[c]]", return "accaccacc".
    s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

    Solution 1: Use 2 stacks
    class Solution:
        def decodeString(self, s: str) -> str:
            char_stack = []
            num_stack = []
            res, num, index = '', 0, 0
            while index < len(s):
                char = s[index]
                if char.isdigit():
                    num = 0
                    while index < len(s) and s[index].isdigit():
                        num = 10 * num + int(s[index])
                        index += 1
                    num_stack.append(num)
                elif char == '[':
                    char_stack.append(res)
                    res = ''
                    index += 1
                elif char == ']':
                    cur_char_lst = list(char_stack.pop())      
                    times = num_stack.pop()
                    for _ in range(times):
                        cur_char_lst.append(res)
                    res = ''.join(cur_char_lst)
                    index += 1
                else:
                    res += char
                    index += 1
            return res

    solution 2: Use 1 stack

    class Solution:
        def decodeString(self, s: str) -> str:
            stack = []
            cur_string, cur_num = '', 0
            for c in s:
                if c.isdigit():
                    cur_num = 10 * cur_num + int(c)
                elif c == '[':
                    stack.append(cur_string)
                    stack.append(cur_num)
                    cur_string = ''
                    cur_num = 0
                elif c == ']':
                    prev_num = stack.pop()
                    prev_string = stack.pop()
                    cur_string = prev_string + prev_num * cur_string
                else:
                    cur_string += c
            return cur_string
    class Solution {
        public String decodeString(String s) {
            char[] charArr = s.toCharArray();
            LinkedList<Object> stack = new LinkedList<>();
            int num = 0;
            for (char ch: charArr) {
                if (Character.isDigit(ch)) {
                    num = 10 * num + (ch - '0');
                } else if (ch == '[') {
                    stack.offerFirst(num);
                    num = 0;
                } else if (ch == ']') {
                    String cur = popBack(stack);
                    Integer preNum = (Integer)stack.pop();
                    for (int i = 0; i < preNum; i++) {
                        stack.offerFirst(cur);
                    }
                } else {
                    stack.offerFirst(String.valueOf(ch));
                }
            }
            return popBack(stack);
        }
        
        private String popBack(LinkedList<Object> stack) {
            LinkedList<String> buffer = new LinkedList<>();
            while (!stack.isEmpty() && (stack.peekFirst() instanceof String)) { 
                buffer.offerFirst((String)stack.pollFirst());
            }
            StringBuilder sb = new StringBuilder();
            while (!buffer.isEmpty()) {
                sb.append(buffer.pollFirst());
            }    
            return sb.toString();     
        }
    }
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  • 原文地址:https://www.cnblogs.com/xuanlu/p/11797621.html
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