198. House Robber（C++实现）

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 andit 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 tonightwithout alerting the police.

解题思路一：动态规划求解，设n-2座房子最多能抢到f(n-2)，n-1座房子能最多抢到f(n-1)，n座房子最多能抢f(n)，最后一座房子里有nMoney，

则可以推导出他们之间的关系（分两种情况，第一种不抢最后一座房子得到收益f(n)=f(n-1)；第二种抢最后一座房子得到收益是f(n)=f(n-2)+nMoney；显然抢钱者会选择最大收益！即f(n)=max{f(n-1),f(n-2)+nMoney}.

程序：

class Solution {
#define _MAX(a,b) ((a)>(b))?(a):(b)
public:
int rob(vector<int>& nums) {//动态规划求解。设n座房子最大抢钱f(n),n-1座房子最大抢f(n-1),则f(n)=max{f(n-1),f(n-2)+nums
}
int n=nums.size();
if(n==0)return 0;
if(n==1)return nums[0];
int d
={0};
d[0]=nums[0];
d[1]=_MAX(nums[0],nums[1]);
for(int i=2;i<n;++i){
d[i]=_MAX(d[i-2]+nums[i],d[i-1]);
}
return d[n-1];
}
};

或者：

class Solution {
#define _MAX(a,b) ((a)>(b))?(a):(b)
public:
int rob(vector<int>& nums) {//动态规划求解。设n座房子最大抢钱f(n),n-1座房子最大抢f(n-1),则f(n)=max{f(n-1),f(n-2)+nums
}
int n=nums.size();
int maxProfit=0;
int take=0;
int nontake=0;
for(int i=0;i<n;++i){
take=nontake+nums[i];
nontake=maxProfit;
maxProfit=_MAX(take,nontake);
}
return maxProfit;
}
};

或者：

class Solution {
public:
int rob(vector<int>& nums) {
int n=nums.size();
if(n==0)return 0;
if(n==1)return nums[0];
int max1=nums[0];
int max2=max(nums[0],nums[1]);
for(int i=2;i<n;i++)
{
int temp=max2;
max2=max(max1+nums[i],max2);
max1=temp;
}
return max2;
}
};