[Swift]LeetCode462. 最少移动次数使数组元素相等 II | Minimum Moves to Equal Array Elements II

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/9792341.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 non-empty integer array, find the minimum number of moves required to make all array elements equal, where a move is incrementing a selected element by 1 or decrementing a selected element by 1.

You may assume the array's length is at most 10,000.

Example:

Input:
[1,2,3]

Output:
2

Explanation:
Only two moves are needed (remember each move increments or decrements one element):

[1,2,3]  =>  [2,2,3]  =>  [2,2,2]

---

给定一个非空整数数组，找到使所有数组元素相等所需的最小移动数，其中每次移动可将选定的一个元素加1或减1。 您可以假设数组的长度最多为10000。

例如:

输入:
[1,2,3]

输出:
2

说明：
只有两个动作是必要的（记得每一步仅可使其中一个元素加1或减1）： 

[1,2,3]  =>  [2,2,3]  =>  [2,2,2]

---

class Solution {
    func minMoves2(_ nums: [Int]) -> Int {
        //转换为变量
        var arr:[Int] = nums
        //对数组进行升序排序
        //sorted只返回排序数组, sort才会修改原数组
        //arr.sort(by: {$1 < $2})
        arr = arr.sorted(by: <)
        let len = nums.count
        var res:Int = 0, mid:Int = arr[len / 2]
        for num in arr
        {
            res += abs(num - mid)
        }
        return res
    }
}

---

24ms

class Solution {
    func minMoves2(_ nums: [Int]) -> Int {
        let a = nums.sorted(by:{ $0 < $1 })
        var ret = 0 ,i = 0 ,j = nums.count - 1
        while i < j {
            ret += a[j] - a[i]
            j -= 1 ; i += 1
        }
        return ret
    }
}

---

28ms

class Solution {
    func minMoves2(_ nums: [Int]) -> Int {        
        var i = 0
        var j = nums.count - 1
        var result = 0
        
        let sortedNums = nums.sorted(by:{ $0 < $1 })
        
        while i < j {
            result += sortedNums[j] - sortedNums[i]
            i += 1
            j -= 1
        }
        
        return result
    }
}

---

32ms

class Solution {
    func minMoves2(_ nums: [Int]) -> Int {
        
        var _num = nums.sorted(by:<)
        
        var middleCount = _num.count/2
        
        var value = _num[middleCount]
        var steps = 0
        for n in _num
        {
            if n != value
            {
               steps += abs(value - n)
            }
            
        }
        
        return steps
    }
}

---

40ms

class Solution 
{
    func minMoves2(_ nums: [Int]) -> Int 
    {
      
        // option 2
        let arr = nums.sorted()
        var i = 0, j = arr.count - 1
        var count = 0
        while i < j
        {
            count += (arr[j] - arr[i])
            i += 1
            j -= 1
        }
        return count
    }
}

---

44ms

class Solution {
    func minMoves2(_ nums: [Int]) -> Int {
        if (nums.count <= 1) {
            return 0
        }
        
        var nums = nums
        nums.sort()
        
        var count = 0
        for i in 0..<nums.count {
            count += abs(nums[nums.count/2] - nums[i])
        }
        return count
        
    }
}

---

48ms

class Solution {
    func minMoves2(_ nums: [Int]) -> Int {
        var raw = nums
        raw.sort { (num1, num2) -> Bool in
            return num1 < num2
        }
        var sum: Int = 0
        let mean = raw[raw.count / 2]
        for i in 0 ..< nums.count {
            if i < raw.count / 2 {
                sum += (mean - raw[i])
            }else {
                sum += (raw[i] - mean)
            }
        }
        return sum
    }
}