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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. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.git
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.github
Example 1:数组
Input: [2,3,2] Output: 3 Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.
Example 2:微信
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.
你是一个专业的小偷,计划偷窃沿街的房屋,每间房内都藏有必定的现金。这个地方全部的房屋都围成一圈,这意味着第一个房屋和最后一个房屋是紧挨着的。同时,相邻的房屋装有相互连通的防盗系统,若是两间相邻的房屋在同一夜被小偷闯入,系统会自动报警。this
给定一个表明每一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状况下,可以偷窃到的最高金额。spa
示例 1:code
输入: [2,3,2] 输出: 3 解释: 你不能先偷窃 1 号房屋(金额 = 2),而后偷窃 3 号房屋(金额 = 2), 由于他们是相邻的。
示例 2:htm
输入: [1,2,3,1] 输出: 4 解释: 你能够先偷窃 1 号房屋(金额 = 1),而后偷窃 3 号房屋(金额 = 3)。 偷窃到的最高金额 = 1 + 3 = 4 。
8ms
1 class Solution { 2 func rob(_ nums: [Int]) -> Int { 3 guard nums.count != 1 else { 4 return nums[0] 5 } 6 7 return max(helper(nums, 0, nums.count - 2), helper(nums, 1, nums.count - 1)) 8 } 9 10 fileprivate func helper(_ nums: [Int], _ start: Int, _ end: Int) -> Int { 11 if start > end { 12 return 0 13 } 14 15 var prev = 0, current = 0 16 17 for i in start...end { 18 (current, prev) = (max(prev + nums[i], current), current) 19 } 20 21 return current 22 } 23 }
8msblog
1 class Solution { 2 3 func rob(_ nums: [Int]) -> Int { 4 5 if nums.count == 0 { return 0 } 6 if nums.count == 1 { return nums[0] } 7 8 return max(rob_nonCircular(nums, start: 0, end: nums.count - 1), rob_nonCircular(nums, start: 1, end: nums.count)) 9 } 10 11 func rob_nonCircular(_ nums: [Int], start: Int, end: Int) -> Int { 12 var a = nums[start], na = 0 // adjacent, non-adjacent 13 14 for i in (start+1)..<end { 15 let tmp = a 16 a = na + nums[i] 17 na = max(na, tmp) 18 } 19 20 return max(a, na) 21 } 22 }
12ms
1 class Solution { 2 var cache: [String: Int] = [:] 3 func rob(_ nums: [Int]) -> Int { 4 return max(rob(nums, startingAt: 1, endingAt: nums.count-1), rob(nums, startingAt: 0, endingAt: nums.count-2)) 5 } 6 7 func rob(_ nums: [Int], startingAt: Int, endingAt: Int) -> Int { 8 if nums.count == 0 { 9 return 0 10 } 11 12 if nums.count == 1 { 13 return nums[0] 14 } 15 if startingAt == endingAt { 16 return nums[startingAt] 17 } 18 19 if endingAt == startingAt+1 { 20 return max(nums[startingAt], nums[endingAt]) 21 } 22 23 let first: Int 24 if let firstValue = cache[key(forStartingAt: startingAt, endingAt: endingAt-2)] { 25 first = firstValue 26 } else { 27 first = rob(nums, startingAt: startingAt, endingAt: endingAt-2) 28 cache[key(forStartingAt: startingAt, endingAt: endingAt-2)] = first 29 } 30 31 let second: Int 32 if let secondValue = cache[key(forStartingAt: startingAt, endingAt: endingAt-1)] { 33 second = secondValue 34 } else { 35 second = rob(nums, startingAt: startingAt, endingAt: endingAt-1) 36 cache[key(forStartingAt: startingAt, endingAt: endingAt-1)] = second 37 } 38 39 return max(nums[endingAt]+first, second) 40 41 } 42 43 func key(forStartingAt startingAt: Int, endingAt: Int) -> String { 44 return "StartingAt:" + String(startingAt) + "Ending At:" + String(endingAt) 45 } 46 47 }
20ms
1 class Solution { 2 func rob(_ nums: [Int]) -> Int { 3 if nums.isEmpty { 4 return 0 5 }else if nums.count <= 3 { 6 return nums.max()! 7 } 8 9 return max(rob(nums, 0, nums.count-1), rob(nums, 1, nums.count)) 10 } 11 12 func rob(_ nums : [Int], _ start : Int, _ end : Int) -> Int { 13 var currMax = nums[start] 14 var nextMax = max(currMax, nums[start+1]) 15 16 for i in start+2..<end { 17 let tmp = nextMax 18 nextMax = max(currMax + nums[i], tmp) 19 currMax = tmp 20 } 21 return nextMax 22 } 23 }
24ms
1 class Solution { 2 func rob(_ nums: [Int]) -> Int { 3 let n = nums.count 4 if n == 0 { return 0 } 5 if n == 1 { return nums[0] } 6 if n == 2 { return max(nums[0], nums[1]) } 7 8 // 不考虑首尾相连,且偷第一个的状况 9 var dp1 = [Int](repeating: 0, count: nums.count + 1) 10 // 不考虑首尾相连,且不偷第一个的状况 11 var dp2 = [Int](repeating: 0, count: nums.count + 1) 12 13 var result = [Int](repeating: 0, count: nums.count + 1) 14 15 dp1[0] = nums[0] 16 dp1[1] = nums[0] 17 for i in 2 ..< nums.count { 18 dp1[i] = max(dp1[i-1], dp1[i-2] + nums[i]) 19 } 20 21 dp2[1] = nums[1] 22 for i in 2 ..< nums.count { 23 dp2[i] = max(dp2[i-1], dp2[i-2] + nums[i]) 24 } 25 26 result[0] = nums[0] 27 for i in 1 ..< nums.count { 28 result[i] = max(dp1[i-1], dp2[i]) 29 } 30 return result[nums.count - 1] 31 32 } 33 }