[Swift]LeetCode927. 三等分 | Three Equal Parts

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/9826329.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given an array A of 0s and 1s, divide the array into 3 non-empty parts such that all of these parts represent the same binary value.

If it is possible, return any [i, j] with i+1 < j, such that:

• A[0], A[1], ..., A[i] is the first part;
• A[i+1], A[i+2], ..., A[j-1] is the second part, and
• A[j], A[j+1], ..., A[A.length - 1] is the third part.
• All three parts have equal binary value.

If it is not possible, return [-1, -1].

Note that the entire part is used when considering what binary value it represents.  For example, [1,1,0] represents 6 in decimal, not 3.  Also, leading zeros are allowed, so [0,1,1] and [1,1] represent the same value.

 Example 1:

Input: [1,0,1,0,1]
Output: [0,3]

Example 2:

Input: [1,1,0,1,1]
Output: [-1,-1]

Note:

1. 3 <= A.length <= 30000
2. A[i] == 0 or A[i] == 1

---

 给定一个由 0 和 1 组成的数组 A，将数组分成 3 个非空的部分，使得所有这些部分表示相同的二进制值。

如果可以做到，请返回任何 [i, j]，其中 i+1 < j，这样一来：

• A[0], A[1], ..., A[i] 组成第一部分；
• A[i+1], A[i+2], ..., A[j-1] 作为第二部分；
• A[j], A[j+1], ..., A[A.length - 1] 是第三部分。
• 这三个部分所表示的二进制值相等。

如果无法做到，就返回 [-1, -1]。

注意，在考虑每个部分所表示的二进制时，应当将其看作一个整体。例如，[1,1,0] 表示十进制中的 6，而不会是 3。此外，前导零也是被允许的，所以 [0,1,1] 和 [1,1] 表示相同的值。

 示例 1：

输入：[1,0,1,0,1]
输出：[0,3]

示例 2：

输出：[1,1,0,1,1]
输出：[-1,-1]

 提示：

1. 3 <= A.length <= 30000
2. A[i] == 0 或 A[i] == 1

---

388ms

class Solution {
  func threeEqualParts(_ A: [Int]) -> [Int] {
    var ones = [Int]()
    for var i in 0..<A.count {
      if A[i] == 1 {
        ones.append(i)
      }
    }    
    if ones.count%3 != 0 {
      return [-1, -1]
    }
    
    if ones.count == 0 {
      return [0, A.count-1]
    }

    let part = ones.count/3
    
    let trZero = A.count - ones.last! - 1
    let i = ones[part-1] + trZero
    let j = ones[part*2-1] + trZero
    
    if i >= ones[part] {
      return [-1, -1]
    }
    if j >= ones[part*2] {
      return [-1, -1]
    }
    
    for var k in 0..<part {
      let a = i - ones[k]
      let b = j - ones[k+part]
      let c = A.count - ones[k+part*2]-1
      if a != b {
        return [-1, -1]
      }
      if b != c {
        return [-1, -1]
      }
    }
    
    return [i, j+1]
  }
}

---

424ms

class Solution {
    func threeEqualParts(_ A: [Int]) -> [Int] {
        let countA:Int = A.count
        var one:Int = 0
        for x in A {one += x}
        if one % 3 != 0 {return [-1,-1]}
        if one == 0 {return [0,countA - 1]}
        one /= 3
        var cc:Int = 0
        var pos:[Int] = [Int](repeating: -2,count: 3)
        var idx:Int = 0
        for i in 0..<countA
        {
            if A[i] == 1 && cc % one == 0
            {                
                pos[idx] = i
                idx += 1       
            }
            cc += A[i]
        }
        var len:Int = countA - pos[2]
        if pos[1] < (pos[0] + len) || pos[2] < (pos[1] + len) {return [-1,-1]}
        var i:Int = pos[0], j:Int = pos[1], k:Int = pos[2]
        repeat
        {
            if (A[i] != A[j] || A[i] != A[k]) {return [-1,-1]};
            i += 1
            j += 1
            k += 1
        }while(k < countA)
        return [pos[0] + len - 1, pos[1] + len]      
    }
}

---

428ms

class Solution {
    func threeEqualParts(_ A: [Int]) -> [Int] {
        let countA:Int = A.count
        var one:Int = 0
        for x in A {one += x}
        if one % 3 != 0 {return [-1,-1]}
        if one == 0 {return [0,countA - 1]}
        one /= 3
        var cc:Int = 0
        var pos:[Int] = [Int](repeating: -2,count: 3)
        var idx:Int = 0
        for i in 0..<countA
        {
            if A[i] == 1 && cc % one == 0
            {                
                pos[idx] = i
                idx += 1       
            }
            cc += A[i]
        }
        var len:Int = countA - pos[2]
        if pos[1] < (pos[0] + len) || pos[2] < (pos[1] + len) {return [-1,-1]}
        var i:Int = pos[0], j:Int = pos[1], k:Int = pos[2]
        repeat
        {
            if (A[i] != A[j] || A[i] != A[k]) {return [-1,-1]};
            i += 1
            j += 1
            k += 1
        }while(k < countA)
        return [pos[0] + len - 1, pos[1] + len]      
    }
}