LeetCode----House Robber

House Robber

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.

Credits:
Special thanks to @ifanchu for adding this problem and creating all test cases. Also thanks to @ts for adding additional test cases.

分析：

题意，给定一个所有元素非负数组，从数组中选择元素，要求在不能选择adjacent的元素的情况下使选择的元素总和最大。

这是一道简单的动态规划题，使用f[i]表示选择到下标为i时已选择元素的最大值，nums为给定数组，其状态转移方程为：

f[i] = max{f[i - 1], f[i - 2] + nums[i]}

代码：

class Solution(object):    def rob(self, nums):        """        :type nums: List[int]        :rtype: int        """        f = [0] * 100  # record the max money        nums_len = len(nums)        if nums_len == 0:            return 0        elif nums_len == 1:            return nums[0]        elif nums_len == 2:            return nums[0] if nums[0] > nums[1] else nums[1]        else:            f[0] = nums[0]            f[1] = nums[0] if nums[0] > nums[1] else nums[1]            for i in range(2, nums_len):                f[i] = max(f[i - 1], f[i - 2] + nums[i])            return f[nums_len - 1]

优化代码1：

class Solution(object):    def rob(self, nums):        """        :type nums: List[int]        :rtype: int        """        nums_len = len(nums)        it1, it2 = 0, 0        for i in range(nums_len):            it1 = it2 if it2 > (it1 + nums[i]) else (it1 + nums[i])             it1, it2 = it2, it1        return it2