[Swift]LeetCode1089. 复写零 | Duplicate Zeros

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/11014391.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 a fixed length array arr of integers, duplicate each occurrence of zero, shifting the remaining elements to the right.

Note that elements beyond the length of the original array are not written.

Do the above modifications to the input array in place, do not return anything from your function. 

Example 1:

Input: [1,0,2,3,0,4,5,0]
Output: null
Explanation: After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]

Example 2:

Input: [1,2,3]
Output: null
Explanation: After calling your function, the input array is modified to: [1,2,3] 

Note:

1. 1 <= arr.length <= 10000
2. 0 <= arr[i] <= 9

---

给你一个长度固定的整数数组 arr，请你将该数组中出现的每个零都复写一遍，并将其余的元素向右平移。

注意：请不要在超过该数组长度的位置写入元素。

要求：请对输入的数组 就地 进行上述修改，不要从函数返回任何东西。 

示例 1：

输入：[1,0,2,3,0,4,5,0]
输出：null
解释：调用函数后，输入的数组将被修改为：[1,0,0,2,3,0,0,4]

示例 2：

输入：[1,2,3]
输出：null
解释：调用函数后，输入的数组将被修改为：[1,2,3] 

提示：

1. 1 <= arr.length <= 10000
2. 0 <= arr[i] <= 9

---

Runtime: 40 ms

Memory Usage: 21.1 MB

class Solution {
    func duplicateZeros(_ arr: inout [Int]) {
        var n:Int = arr.count
        var a:[Int] = arr
        var p:Int = 0
        for i in 0..<n
        {
            if a[i] == 0
            {
                if p < n 
                {
                    arr[p] = 0
                    p += 1
                }
                if p < n
                {
                    arr[p] = 0
                    p += 1
                }
            }
            else
            {
                if p < n
                {
                    arr[p] = a[i]
                    p += 1
                }
            }
        }
        
    }
}

---

44ms

class Solution {
    func duplicateZeros(_ arr: inout [Int]) {
        let c = arr.count
        if c <= 1 { return }

        var i = 0
        while i < c {
            if arr[i] == 0 {
                arr.insert(0, at: i)
                i += 2
            } else {
                i += 1
            }
        }
        while arr.count > c {
            _ = arr.popLast()
        }
    }
}

---

48ms

class Solution {
    func duplicateZeros(_ arr: inout [Int]) {
        var count = 0
        for (index, value) in arr.enumerated() {
            if value == 0 {
                count += 1
            }
        }
        var index = arr.count - 1
        while index >= 0 {
            let value = arr[index]
            if index + count < arr.count {
                arr[index + count] = value
            }
            if value == 0 {
                count -= 1
                if index + count < arr.count {
                    arr[index + count] = 0
                }
            }
            index -= 1
        }
    }
}