【leetcode】1209. Remove All Adjacent Duplicates in String II

题目如下：

Given a string s, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them causing the left and the right side of the deleted substring to concatenate together.

We repeatedly make k duplicate removals on s until we no longer can.

Return the final string after all such duplicate removals have been made.

It is guaranteed that the answer is unique.

Example 1:

Input: s = "abcd", k = 2
Output: "abcd"
Explanation: There‘s nothing to delete.

Example 2:

Input: s = "deeedbbcccbdaa", k = 3
Output: "aa"
Explanation: 
First delete "eee" and "ccc", get "ddbbbdaa"
Then delete "bbb", get "dddaa"
Finally delete "ddd", get "aa"

Example 3:

Input: s = "pbbcggttciiippooaais", k = 2
Output: "ps" 

Constraints:

• 1 <= s.length <= 10^5
• 2 <= k <= 10^4
• s only contains lower case English letters.

解题思路：本题解法不难，利用入栈出栈的思路即可。但有几点注意一下，一是只有长度为k的连续出现的相同的字符才能消除，如果有(k+1)个字符a的话，只能消除k个，留下剩余的一个a；同时注意消除后的相同字符的合并。

代码如下：

class Solution(object):
    def removeDuplicates(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: str
        """
        stack = []
        s += ‘#‘
        last_char = None
        continuous = 1
        for i in s:
            if last_char == None:
                last_char = i
            elif last_char == i:
                continuous += 1
            else:
                stack.append([last_char,continuous])
                last_char = i
                continuous = 1
        #print stack
        for i in range(len(stack)-1,-1,-1):
            if stack[i][1] >= k:
                if stack[i][1] % k == 0:
                    del stack[i]
                else:
                    stack[i][1] = stack[i][1] % k
            if i < len(stack) - 1 and stack[i][0] == stack[i+1][0]:
                stack[i][1] += stack[i+1][1]
                del stack[i+1]
            if i < len(stack) and stack[i][1] >= k:
                if stack[i][1] % k == 0:
                    del stack[i]
                else:
                    stack[i][1] = stack[i][1] % k
        res = ‘‘
        for char,count in stack:
            res += char*count
        return res