198. House Robber-动态规划

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.

tags: dp/动态规划

比如：     

 nums:   3      1      5      2      0      7      4

                 dp    :   3      3      ?

求dp[2]：由于dp[1]=dp[0] ，说明在求dp[1]时没有用到nums[1]，所以，不论nums[2]多大，直接加上dp[1]或dp[0] ，dp[2]  =  dp[1] + nums[2]。

                 nums :   3      1      5      2      0      7      4

                 dp     :   3      3      8      ?

求dp[3]：由于dp[2]>dp[1] ，说明在求dp[2]时用到了nums[2]，因为nums[3]不能和nums[2]同时出现，所以要判断dp[1]+nums[3]和dp[2]的大小。dp[3] = max(dp[1]+nums[3],dp[2])

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