[Swift]LeetCode1029. 两地调度 | Two City Scheduling

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There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that exactly Npeople arrive in each city.

Example 1:

Input: [[10,20],[30,200],[400,50],[30,20]]
Output: 110
Explanation: 
The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.

The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.

Note:

1. 1 <= costs.length <= 100
2. It is guaranteed that costs.length is even.
3. 1 <= costs[i][0], costs[i][1] <= 1000

---

公司计划面试 2N 人。第 i 人飞往 A 市的费用为 costs[i][0]，飞往 B 市的费用为 costs[i][1]。

返回将每个人都飞到某座城市的最低费用，要求每个城市都有 N 人抵达。

示例：

输入：[[10,20],[30,200],[400,50],[30,20]]
输出：110
解释：
第一个人去 A 市，费用为 10。
第二个人去 A 市，费用为 30。
第三个人去 B 市，费用为 50。
第四个人去 B 市，费用为 20。

最低总费用为 10 + 30 + 50 + 20 = 110，每个城市都有一半的人在面试。

提示：

1. 1 <= costs.length <= 100
2. costs.length 为偶数
3. 1 <= costs[i][0], costs[i][1] <= 1000

---

Runtime: 16 ms

Memory Usage: 19.3 MB

class Solution {
    func twoCitySchedCost(_ costs: [[Int]]) -> Int {
        var base:Int = 0
        var n:Int = costs.count
        var cs:[Int] = [Int](repeating:0,count:n)
        for (index,cost) in costs.enumerated()
        {
            base += cost[0]
            cs[index] = cost[1] - cost[0]  
        }
        cs.sort()
        for i in 0..<n/2
        {
            base += cs[i]
        }
        return base
    }
}

---

16ms

class Solution {
    func twoCitySchedCost(_ costs: [[Int]]) -> Int {
        var costDiff = costs.sorted(by: { $0[1] - $0[0]  > $1[1] - $1[0] })
        var minCost = 0
        for i in 0..<costs.count {
            if i < costs.count / 2 {
                minCost += costDiff[i][0]
            } else {
                minCost += costDiff[i][1]
            }
        }
        return minCost
    }
}

---

20ms

class Solution {
    func twoCitySchedCost(_ costs: [[Int]]) -> Int {           
        let costs = costs.sorted { (a, b) in
            return abs(a[0] - a[1]) > abs(b[0] - b[1])
        }
        let N = costs.count / 2
        var c1 = 0, c2 = 0, ans = 0
        for i in 0..<2*N {
            if ((costs[i][0] < costs[i][1] && c1 < N) || c2 == N) {
                ans += costs[i][0]
                c1 += 1
            }
            else {
                ans += costs[i][1]
                c2 += 1
            }
        }
        return ans

    }
}

---

24ms

class Solution {
    func twoCitySchedCost(_ costs: [[Int]]) -> Int {
        
        var count = 0
        var deltaAtoB = [Int]()
        var deltaBtoA = [Int]()
        var result = 0
        for cost in costs {
            if cost[0] <= cost[1] {
                count += 1
                result += cost[0]
                deltaBtoA.append(cost[1] - cost[0])
            } else {
                count -= 1
                result += cost[1]
                deltaAtoB.append(cost[0] - cost[1])
            }
        }

        if count == 0 {
            return result
        } else if count > 0 {
            deltaBtoA.sort()
            for i in 0..<count/2 {
                result += deltaBtoA[i]
            }
        } else {
            deltaAtoB.sort()
            for i in 0..<abs(count)/2 {
                result += deltaAtoB[i]
            }
        }
        return result
    }
}

---

28ms

class Solution {
    func twoCitySchedCost(_ costs: [[Int]]) -> Int {
        var costDiff = costs.sorted(by: { $0[1] - $0[0]  > $1[1] - $1[0] })
        var minCost = 0
        for i in 0..<costs.count {
            if i < costs.count / 2 {
                minCost += costDiff[i][0]
            } else {
                minCost += costDiff[i][1]
            }
        }
        return minCost
    }
}

---

32ms

class Solution {
    func twoCitySchedCost(_ costs: [[Int]]) -> Int {
        let N = costs.count / 2
        var dp = [[Int]]()
        
        for i in 0...N {
            let row = Array(repeating: 0, count: N+1)
            dp.append(row)
        }
        
        for i in 1...N {
            dp[i][0] = dp[i-1][0] + costs[i-1][0]
        }
        
        for j in 1...N {
            dp[0][j] = dp[0][j-1] + costs[j-1][1]
        }
        
        for i in 1...N {
            for j in 1...N {
                dp[i][j] = min(dp[i-1][j] + costs[i+j-1][0], dp[i][j-1] + costs[i+j-1][1])
            }
        }
        
        return dp[N][N]
    }
}

---

44ms

class Solution {
    func twoCitySchedCost(_ costs: [[Int]]) -> Int {
        var costsrt = costs.sorted(by: {return abs($0[0]-$0[1]) >= abs($1[0]-$1[1])})
        
        var N = costs.count / 2
        var A = 0
        var B = 0
        var ac = 0
        var bc = 0
        
        for p in costsrt {
            if (p[0] > p[1] && bc < N) || ac == N {
                B += p[1]
                bc += 1
                print("B:", p[1])
            } else {
                A += p[0]
                ac += 1
                print("A:", p[0])
            }
        }
        
        return A+B
    }
}