[Swift]LeetCode696. 計數二進制子串 | Count Binary Substrings

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號：山青詠芝（shanqingyongzhi）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址： http://www.javashuo.com/article/p-mfjwxcij-me.html 
➤若是連接不是山青詠芝的博客園地址，則多是爬取做者的文章。
➤原文已修改更新！強烈建議點擊原文地址閱讀！支持做者！支持原創！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Give a string s, count the number of non-empty (contiguous) substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively.

Substrings that occur multiple times are counted the number of times they occur.

Example 1:

Input: "00110011"
Output: 6
Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".

Notice that some of these substrings repeat and are counted the number of times they occur.

Also, "00110011" is not a valid substring because all the 0's (and 1's) are not grouped together.

Example 2:

Input: "10101"
Output: 4
Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.

Note:

• s.length will be between 1 and 50,000.
• s will only consist of "0" or "1" characters.

---

給定一個字符串 s，計算具備相同數量0和1的非空(連續)子字符串的數量，而且這些子字符串中的全部0和全部1都是組合在一塊兒的。

重複出現的子串要計算它們出現的次數。

示例 1 :

輸入: "00110011"
輸出: 6
解釋: 有6個子串具備相同數量的連續1和0：「0011」，「01」，「1100」，「10」，「0011」 和 「01」。

請注意，一些重複出現的子串要計算它們出現的次數。

另外，「00110011」不是有效的子串，由於全部的0（和1）沒有組合在一塊兒。

示例 2 :

輸入: "10101"
輸出: 4
解釋: 有4個子串：「10」，「01」，「10」，「01」，它們具備相同數量的連續1和0。

注意：

• s.length 在1到50,000之間。
• s 只包含「0」或「1」字符。

---

80ms

 1 class Solution {
 2     func countBinarySubstrings(_ s: String) -> Int {
 3         var chars = Array(s)
 4         var prev = 0, cur = 1
 5         var result = 0
 6         for i in 1..<chars.count {
 7             if chars[i-1] != chars[i] {
 8                 result += min(prev, cur)
 9                 prev = cur
10                 cur = 1
11                 
12             } else{
13                 cur += 1
14             }
15         }
16         result += min(prev, cur)
17         return result
18     }
19 }

---

Runtime: 92 ms

Memory Usage: 19.4 MB

 1 class Solution {
 2     func countBinarySubstrings(_ s: String) -> Int {
 3         var arr:[Character] = Array(s)
 4         var res:Int = 0
 5         var pre:Int = 0
 6         var cur:Int = 1
 7         var n:Int = s.count
 8         for i in 1..<n
 9         {
10             if arr[i] == arr[i - 1]
11             {
12                 cur += 1              
13             }
14             else
15             {
16                 pre = cur
17                 cur = 1
18             }
19             if pre >= cur
20             {
21                 res += 1
22             }
23         }
24         return res
25     }
26 }

---

112ms

 1 class Solution {
 2   func countBinarySubstrings(_ s: String) -> Int {
 3     var zero = 0
 4     var one = 0
 5     
 6     var substrCount = 0
 7     
 8     let s = Array(s)
 9     
10     var i = 0
11     while i < s.count {
12       while i < s.count && s[i] == "0" {
13         zero += 1
14         if one > 0 {
15           substrCount += 1
16           one -= 1
17         }
18         i += 1
19       }
20       
21       one = 0
22       
23       while i < s.count && s[i] == "1" {
24         one += 1
25         if zero > 0 {
26           substrCount += 1
27           zero -= 1
28         }
29         i += 1
30       }
31       
32       zero = 0
33     }
34     
35     return substrCount
36   }
37 }

---

136ms

 1 class Solution {
 2     func countBinarySubstrings(_ s: String) -> Int {
 3         var chs = Array(s)
 4         
 5         var groups:[Int] = [Int]()
 6         
 7         groups.append(1)
 8         
 9         for i in 1..<chs.count {
10             if chs[i] == chs[i - 1] {
11                 groups[groups.count - 1] += 1
12             } else {
13                 groups.append(1)
14             }
15         }
16         
17         var res = 0
18         for i in 1..<groups.count {
19             res += min(groups[i], groups[i - 1])
20         }
21         
22         return res
23     }
24 }

---

148ms

 1 class Solution {
 2     func countBinarySubstrings(_ s: String) -> Int {
 3         var zeros = 0
 4         var ones = 0
 5         var res = 0
 6         var isOne = s.first! == "1"
 7         for char in s {
 8             if char == "1" {
 9                 if !isOne { ones = 0 }
10                 isOne = true
11                 ones += 1
12                 if zeros > 0 {
13                     res += 1
14                     zeros -= 1
15                 }
16             } else {
17                 if isOne { zeros = 0 }
18                 isOne = false
19                 
20                 zeros += 1
21                 if ones > 0 {
22                     res += 1
23                     ones -= 1
24                 }
25             }
26         }
27         return res
28     }
29 }

---

188ms

 1 class Solution {
 2     func countBinarySubstrings(_ s: String) -> Int {
 3         guard s.count > 1 else { return 0 }        
 4         var sArray = s.map{ String($0) }        
 5         var count = 0
 6         for i in 0..<sArray.count-1 {            
 7             if sArray[i] != sArray[i+1] {
 8                 count += findPalindrom(at: i, in: sArray)
 9             }            
10         }        
11         return count        
12     }
13     
14     
15     func findPalindrom(at index: Int, in array: [String])->Int {        
16         guard index < array.count-1 else { return 0 }        
17         var count = 1        
18         var left = index-1
19         var right = index+2        
20         while left >= 0 && right < array.count {
21             if array[left] == array[left+1] && array[right] == array[right-1] { count += 1 }
22             else { break }            
23             left -= 1
24             right += 1
25         }        
26         return count
27     }        
28 }
29