LeetCode OJ 之 House Robber（抢劫犯）

题目：

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.
假设你是一个专业抢劫犯，计划抢劫一条街道的房子。每个房子内都有一定数量的钱，唯一限制你抢劫的是每相邻的两个房子都有安保系统连接。如果两个相邻的房子都被破门而入的话，安保系统会自动报警。
有一组非负整数代表每个房子内的钱数，判断如何在不惊动警察的情况下可以抢劫的最多钱数。

思路：

动态规划：假设 f [ i ] 表示到第 i 个房子抢到的钱，则可得到状态转移方程为：f [ i ] = max( f [ i - 2 ] + nums [ i ] , f [ i - 1 ]).

代码：

class Solution {
public:
int rob(vector<int>& nums)
{
int len = nums.size();
if(len == 0) return 0;
if(len == 1) return nums[0];
if(len == 2) return max(nums[0] , nums[1]);
vector<int> f(len , 0);
f[0] = nums[0];
f[1] = max(nums[0] , nums[1]);
for(int i = 2 ; i < len ; i++)
f[i] = max(f[i-2] + nums[i] , f[i-1]);
return f[len-1];
}
};

优化空间复杂度：
class Solution {
public:
int rob(vector<int>& nums)
{
int len = nums.size();
int f0 = 0 , f1 = 0 , f2 = 0;
for(int i = 0 ; i < len ; i++)
{
f2 = max(f0 + nums[i] , f1);
f0 = f1 ;
f1 = f2;
}
return f2;
}
};