1 """ 2 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. 3 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. 4 Example 1: 5 Input: [1,2,3,1] 6 Output: 4 7 Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). 8 Total amount you can rob = 1 + 3 = 4. 9 Example 2: 10 Input: [2,7,9,3,1] 11 Output: 12 12 Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1). 13 Total amount you can rob = 2 + 9 + 1 = 12. 14 """ 15 """ 16 看到这道题，把问题想简单了 17 思路是 奇数列求和 ，偶数列求和，取最大的值。 18 其实不然。对于下面这个用例是 wrong answer 19 Input 20 [2,1,1,2] 21 Output 22 3 23 Expected 24 4 25 """ 26 class Solution1(object): #Wrong Answer 27 def rob(self, nums): 28 n = len(nums) 29 if n%2 == 0: 30 even, odd = 0, 0 31 for i in range(0, n, 2): 32 even += nums[i] 33 odd += nums[i+1] 34 else: 35 even, odd = nums[0], 0 36 for i in range(1, n, 2): 37 odd += nums[i] 38 even += nums[i+1] 39 return max(even, odd) 40 41 """ 42 下面是正确的思路。动态规划 43 开辟一个dp数组，dp[i]表示从0-i户可以打劫到的最大钱数。 44 则有dp[i] = max(dp[i-1],dp[i-2]+nums[i])。 45 第（i-1）户打劫到的最大钱数+不打劫第i户， 46 与第（i-2）户打劫的最大钱数+打劫第i户， 47 两者中的最大值。 48 还要分情况把nums==None, len(nums)==1, len(nums)==2 分别讨论 49 """ 50 class Solution(object): 51 def rob(self, nums): 52 if nums == []: #bug if nums == None or len(nums) < 0: 53 return 0 54 if len(nums) == 1: #len为1的情况 55 return nums[0] 56 if len(nums) == 2: #len为2的情况 57 return max(nums[0], nums[1]) 58 dp = [0] * len(nums) #开辟dp数组，记录最大的钱数 59 dp[0] = nums[0] #将dp[0]初始化 60 dp[1] = max(nums[0], nums[1]) #将dp[1]初始化 61 for i in range(2, len(nums)): 62 dp[i] = max(dp[i-1], nums[i] + dp[i-2]) #！！！动态转移方程 63 return max(dp)