leetcode#394. Decode String

Description

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".

Code

class Solution:
    def decodeString(self, s):
        """
        :type s: str
        :rtype: str
        """
        res = ''
        dig = 0
        stack = []
        cntable = True
        for index, i in enumerate(s):
            if cntable and '0' <= i and i <= '9':
                dig = dig * 10 + int(i)
            else:
                if i == '[':
                    stack.append(index)
                elif i == ']':
                    start = stack.pop()
                    if stack == []:
                        temp = self.decodeString(s[start + 1:index])
                        res += dig * temp
                        dig = 0
                        cntable = True
                        #return res + self.decodeString(s[index + 1:])也可以这么做
                elif stack == []:
                    res += i
                if not dig == 0:
                    cntable = False
        return res

同样的思路，我觉得下面这位大神的写法值得借鉴：

Code2(ayushpatwari )

class Solution(object):
    def decodeString(self, s):
        stack = []
        stack.append(["", 1])
        num = ""
        for ch in s:
            if ch.isdigit():
              num += ch
            elif ch == '[':
                stack.append(["", int(num)])
                num = ""
            elif ch == ']':
                st, k = stack.pop()
                stack[-1][0] += st*k
            else:
                stack[-1][0] += ch
        return stack[0][0]

Conclusion

同样的思路，我的写法是用递归，而ayushpatwari 这位大神的写法是将’[‘前的状态保存至stack中，然后遇到’]’时pop栈顶，再将最新栈顶的值更新，最后将栈底的值输出。感觉代码更加简洁，优雅。