[Swift]LeetCode198. 打家劫舍 | House Robber

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➤微信公众号：山青咏芝（shanqingyongzhi）
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➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/9744435.html 
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You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

---

你是一个专业的小偷，计划偷窃沿街的房屋。每间房内都藏有一定的现金，影响你偷窃的唯一制约因素就是相邻的房屋装有相互连通的防盗系统，如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警。

给定一个代表每个房屋存放金额的非负整数数组，计算你在不触动警报装置的情况下，能够偷窃到的最高金额。

示例 1:

输入: [1,2,3,1]
输出: 4
解释: 偷窃 1 号房屋 (金额 = 1) ，然后偷窃 3 号房屋 (金额 = 3)。
     偷窃到的最高金额 = 1 + 3 = 4 。

示例 2:

输入: [2,7,9,3,1]
输出: 12
解释: 偷窃 1 号房屋 (金额 = 2), 偷窃 3 号房屋 (金额 = 9)，接着偷窃 5 号房屋 (金额 = 1)。
     偷窃到的最高金额 = 2 + 9 + 1 = 12 。

---

8ms

class Solution {
    func rob(_ nums: [Int]) -> Int {
        
        if nums.count <= 0 {
            return 0
        }
        
        var last: Int = 0
        var now: Int = 0
        
        for i in 0...nums.count - 1 {
            var temp = last
            last = now
            now = max(temp + nums[i],now)
        }
        return now
    }
}

---

8ms

class Solution {
    func rob(_ nums: [Int]) -> Int {
    var mem : [Int?] = [Int?](repeating: nil, count: nums.count)
    return robHouse(nums, index: 0, mem: &mem)
}

func robHouse(_ nums: [Int], index : Int, mem : inout [Int?])->Int{
    guard index != nums.count else{
        return 0
    }
    
    guard index != nums.count - 1 else{
        if let answer = mem[index]{
            return answer
        }else{
            mem[index] = nums[index]
            return mem[index]!
        }
    }
    
    guard index != nums.count - 2 else {
        if let answer = mem[index]{
            return answer
        }else{
            mem[index] = max(nums[index], nums[index + 1])
            return mem[index]!
        }
    }
    if let answer = mem[index]{
        return answer
    }else{
        mem[index] = max(nums[index] + robHouse(nums, index: index + 2, mem: &mem), nums[index + 1] + robHouse(nums, index: index + 3, mem: &mem))
        return mem[index]!
    }
    
}
}

---

12ms

class Solution {
    func rob(_ nums: [Int]) -> Int {
        var dp = [Int]()
        for _ in 0...nums.count {
            dp.append(0)
        }
        for i in 0..<dp.count {
            if i == 0 {
                dp[0] = 0
                continue
            }
            if i == 1 {
                dp[1] = nums[0]
                continue
            }
            dp[i] = max(dp[i - 2] + nums[i - 1], dp[i - 1])
        }
        return dp[nums.count]
    }
}

---

12ms

class Solution {
    func rob(_ nums: [Int]) -> Int {

        var n = nums.count
        if n == 0 { return 0 }
        if n == 1 { return nums[0] }
        var dp = [nums[0], max(nums[0], nums[1])]
        for i in 2..<n {
             dp.append(max(dp[i-1], dp[i-2]+nums[i]))
        }
        return dp.last!
    }
}

---

16ms

class Solution {
    func rob(_ nums: [Int]) -> Int {
        var currentMax = 0
        var previousMax = 0
        var result = 0
        
        for num in nums {
            result = max(currentMax, previousMax + num)
            previousMax = currentMax
            currentMax = result
        }
        return result
    }
}