题目如下:
Given a string
s
, a k duplicate removal consists of choosingk
adjacent and equal letters froms
and removing them causing the left and the right side of the deleted substring to concatenate together.We repeatedly make
k
duplicate removals ons
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
原文地址:https://www.cnblogs.com/seyjs/p/11616707.html
时间: 2024-11-05 13:42:53