[LeetCode]198. 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.

Solution:

DP 解法:

1. 这条街只有1间房: 无疑这间就是打劫对象, dp[0] = num[0]

2. 这条街有两间房: 则选择打劫其中一家钱数较多的, dp[1] = max(num[0], num[1])

3. 这条街有n(n>=3)间房: 打劫到k间时,由于不能打劫相邻的房屋,所获得的最大钱数 dp[k] = max(dp[k-2] + num[k], dp[k-1])

Time Complexity: O(n)

Space: O(n)

Note: 开始时没有看清题意,这道题是从第一间房依次开始打劫,我起初理解为在此街道上随机选取一家开始,惭愧惭愧~

class Solution:
    # @param num, a list of integer
    # @return an integer
    def rob(self, num):
        length = len(num)
        if length == 0:
            return 0
        dp = [0]*length
        dp[0] = num[0]
        if length == 1:
            return num[0]
        dp[1] = max(num[0], num[1])
        for i in range(2, length):
            dp[i] = max(dp[i-1], dp[i-2]+num[i])
        return dp[length-1]


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