[Swift]LeetCode1047.删除字符串中的全部相邻重复项 | Remove All Adjacent Duplicates In String

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Given a string S of lowercase letters, a duplicate removal consists of choosing two adjacent and equal letters, and removing them.

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

Return the final string after all such duplicate removals have been made.  It is guaranteed the answer is unique.

Example 1:

Input: "abbaca"
Output: "ca" Explanation: For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move.  The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca".

Note:

1. 1 <= S.length <= 20000
2. S consists only of English lowercase letters.

---

给出由小写字母组成的字符串 S，重复项删除操做会选择两个相邻且相同的字母，并删除它们。

在 S 上反复执行重复项删除操做，直到没法继续删除。

在完成全部重复项删除操做后返回最终的字符串。答案保证惟一。

示例：

输入："abbaca"
输出："ca"
解释：
例如，在 "abbaca" 中，咱们能够删除 "bb" 因为两字母相邻且相同，这是此时惟一能够执行删除操做的重复项。以后咱们获得字符串 "aaca"，其中又只有 "aa" 能够执行重复项删除操做，因此最后的字符串为 "ca"。

提示：

1. 1 <= S.length <= 20000
2. S 仅由小写英文字母组成。

---

Runtime: 132 ms

Memory Usage: 21.4 MB

 1 class Solution {
 2     func removeDuplicates(_ S: String) -> String {
 3         var res:String = String()
 4         for c in S
 5         {
 6             if res.isEmpty || res.last! != c
 7             {
 8                 res.append(c)
 9             }
10             else
11             {
12                 res.popLast()
13             }
14         }
15         return res        
16     }
17 }

---

132ms

 1 class Solution {
 2   func removeDuplicates(_ S: String) -> String {
 3     var s = S
 4     var i = s.startIndex
 5     while !s.isEmpty && s.index(after: i) != s.endIndex {
 6       if s[i] == s[s.index(after: i)] {
 7         s.remove(at: i)
 8         s.remove(at: i)
 9         if i != s.startIndex {
10           i = s.index(before: i)
11         }
12         continue
13       }
14       i = s.index(after: i)
15     }
16     return s
17   }
18 }

---

144ms

 1 class Solution {
 2     func removeDuplicates(_ S: String) -> String
 3     {
 4         var list: [Character] = []
 5         for c in S
 6         {
 7             if let last = list.last
 8             {
 9                 if last == c {
10                     list.removeLast()
11                 }
12                 else { list.append(c) }
13             }
14             else {
15                 list.append(c)
16             }
17         }
18         return String(list)
19     }
20 }

---

196ms

 1 class Solution {
 2     func removeDuplicates(_ S: String) -> String {
 3         var stack = [Character]()
 4         
 5         for char in S {
 6             if stack.last == char {
 7                 stack.removeLast()
 8             } else {
 9                 stack.append(char)
10             }
11         }
12         
13         return String(stack)
14     }
15 }