[Swift]LeetCode75. 颜色分类 | Sort Colors

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/9929610.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 with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue.

Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

Note: You are not suppose to use the library's sort function for this problem.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

Follow up:

• A rather straight forward solution is a two-pass algorithm using counting sort.
  First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's.
• Could you come up with a one-pass algorithm using only constant space?

---

给定一个包含红色、白色和蓝色，一共 n 个元素的数组，原地对它们进行排序，使得相同颜色的元素相邻，并按照红色、白色、蓝色顺序排列。

此题中，我们使用整数 0、 1 和 2 分别表示红色、白色和蓝色。

注意:
不能使用代码库中的排序函数来解决这道题。

示例:

输入: [2,0,2,1,1,0]
输出: [0,0,1,1,2,2]

进阶：

• 一个直观的解决方案是使用计数排序的两趟扫描算法。
  首先，迭代计算出0、1 和 2 元素的个数，然后按照0、1、2的排序，重写当前数组。
• 你能想出一个仅使用常数空间的一趟扫描算法吗？

---

8ms

class Solution {
    func sortColors(_ nums: inout [Int]) {
        
        var i = 0
        var left = 0
        var right = nums.count - 1
        
        while i <= right {
            if nums[i] == 0 {
                nums.swapAt(i, left)
                i += 1
                left += 1
            } else if nums[i] == 1 {
                i += 1
            } else {
                nums.swapAt(i, right)
                right -= 1
            }
        }
    }
}

---

12ms

class Solution {
    /// 桶排序法, 0...red 都是 0，blue...nums.count-1 都是2
    func sortColors(_ nums: inout [Int]) {
        var red = -1
        var blue = nums.count
        var i = 0
        
        while i < blue {
            if nums[i] == 0 {
                red += 1
                (nums[i], nums[red]) = (nums[red], nums[i])
                i += 1
            } else if nums[i] == 2 {
                blue -= 1
                (nums[i], nums[blue]) = (nums[blue], nums[i])
            } else {
                i += 1
            }
        }
    }
}

---

16ms

class Solution {
    func sortColors(_ nums: inout [Int]) {
        var zeroCounter = 0
        var oneCounter = 0
        var twoCounter = 0
        
        for each in nums{
            if each == 0{
                zeroCounter += 1
            }else if each == 1{
                oneCounter += 1
            }else if each == 2{
                twoCounter += 1
            }
        }
        
        for each in 0..<zeroCounter{
            nums[each] = 0
        }
        
        for each in 0..<oneCounter{
            nums[zeroCounter+each] = 1
        }
        
        for each in 0..<twoCounter{
            nums[zeroCounter+oneCounter+each] = 2
        }
        
    }
}

---

20ms

class Solution {
    func sortColors(_ nums: inout [Int]) {
            var redCount = 0, whiteCount = 0, blueCount = 0
    for i in 0..<nums.count {
        if nums[i] == 0 {
            redCount = redCount + 1
            nums[redCount - 1] = 0
        } else if nums[i] == 1 {
            whiteCount = whiteCount + 1
        } else {
            blueCount = blueCount + 1
        }
    }
    for i in redCount..<redCount+whiteCount {
        nums[i] = 1
    }
    for i in redCount+whiteCount..<nums.count {
        nums[i] = 2
    }
    }
}