[Swift]LeetCode1003. 检查替换后的词是否有效 | Check If Word Is Valid After Substitutions
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We are given that the string "abc" is valid.
From any valid string V, we may split V into two pieces X and Y such that X + Y (X concatenated with Y) is equal to V. (X or Y may be empty.) Then, X + "abc" + Y is also valid.
If for example S = "abc", then examples of valid strings are: "abc", "aabcbc", "abcabc", "abcabcababcc". Examples of invalid strings are: "abccba", "ab", "cababc", "bac".
Return true if and only if the given string S is valid.
Example 1:
Input: true Explanation: We start with the valid string "abc". Then we can insert another "abc" between "a" and "bc", resulting in "a" + "abc" + "bc" which is "aabcbc".
Example 2:
Input: true Explanation: "abcabcabc" is valid after consecutive insertings of "abc". Then we can insert "abc" before the last letter, resulting in "abcabcab" + "abc" + "c" which is "abcabcababcc".
Example 3:
Input: false
Example 4:
Input: false
Note:
1 <= S.length <= 20000S[i]is'a','b', or'c'
给定有效字符串 "abc"。
对于任何有效的字符串 V，我们可以将 V 分成两个部分 X 和 Y，使得 X + Y（X 与 Y 连接）等于 V。（X 或 Y 可以为空。）那么，X + "abc" + Y 也同样是有效的。
例如，如果 S = "abc"，则有效字符串的示例是："abc"，"aabcbc"，"abcabc"，"abcabcababcc"。无效字符串的示例是："abccba"，"ab"，"cababc"，"bac"。
如果给定字符串 S 有效，则返回 true；否则，返回 false。
示例 1：
输入："aabcbc" 输出：true 解释： 从有效字符串 "abc" 开始。 然后我们可以在 "a" 和 "bc" 之间插入另一个 "abc"，产生 "a" + "abc" + "bc"，即 "aabcbc"。
示例 2：
输入："abcabcababcc" 输出：true 解释： "abcabcabc" 是有效的，它可以视作在原串后连续插入 "abc"。 然后我们可以在最后一个字母之前插入 "abc"，产生 "abcabcab" + "abc" + "c"，即 "abcabcababcc"。
示例 3：
输入："abccba" 输出：false
示例 4：
输入："cababc" 输出：false
提示：
1 <= S.length <= 20000S[i]为'a'、'b'、或'c'
44ms
1 class Solution
2 {
3 func isValid( _ string: String ) -> Bool
4 {
5 var aStack: Int = 0
6 var bStack: Int = 0
7 for char in string.characters
8 {
9 if char == "a"
10 {
11 aStack += 1
12 }
13 else if char == "b"
14 {
15 bStack += 1
16 if bStack > aStack
17 {
18 return false
19 }
20 }
21 else if char == "c"
22 {
23 aStack -= 1
24 bStack -= 1
25 if aStack < 0 || bStack < 0
26 {
27 return false
28 }
29 }
30 }
31 return aStack == 0 && bStack == 0
32 }
33 }52ms
1 class Solution {
2 func isValid(_ S: String) -> Bool {
3 var stack = [Character]()
4 for char in S {
5 if char == "c" {
6 guard stack.count >= 2 else { return false }
7 guard stack.removeLast() == "b" else { return false }
8 guard stack.removeLast() == "a" else { return false }
9 } else {
10 stack.append(char)
11 }
12 }
13 return stack.count == 0
14 }
15 }56ms
1 class Solution {
2 func isValid(_ S: String) -> Bool {
3 var stack: [Character] = []
4 for s in S {
5 switch s {
6 case "a", "b": stack.append(s)
7 case "c":
8 guard stack.count >= 2,
9 stack.removeLast() == "b",
10 stack.removeLast() == "a" else {
11 return false
12 }
13 default: fatalError()
14 }
15 }
16
17 return stack.isEmpty
18 }
19 }80ms
1 class Solution {
2 func isValid(_ S: String) -> Bool {
3 let s = Array(S)
4 var v = [Character]()
5 var j = 0
6 while j < s.count {
7 v.append(s[j])
8 if v.count >= 3 &&
9 v[v.count - 3] == "a" &&
10 v[v.count - 2] == "b" &&
11 v[v.count - 1] == "c" {
12 v.removeLast(3)
13 }
14 j += 1
15 }
16 return v.count == 0
17 }
18 }288ms
1 class Solution {
2 func isValid(_ S: String) -> Bool {
3 var S = S
4 while S.contains("abc") {
5 S = S.replacingOccurrences(of: "abc", with: "", options: .literal, range: nil)
6 }
7 return S.isEmpty
8 }
9 }292ms
1 class Solution {
2 func isValid(_ S: String) -> Bool {
3 var curS = S
4 while curS.count >= 3 {
5 let str = (curS as NSString).replacingOccurrences(of: "abc", with: "")
6
7 curS = str
8 if !(str as NSString).contains("abc") { break}
9 }
10 return curS.count == 0
11 }
12 }