LEET code -- 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.

跟之前爬樓梯一樣
我只要把之前算好的拿來用就好

假設房子價錢如下
   1  4  7  1  8  13   6  22  6   8   33 123  55]
我們再去每搶一個房子會得到的價錢weight
weight[i+1]= weight[i+1-2]>= weight[i+1-3]? weight[i+1-2]+nums[i]:weight[i+1-3]+nums[i];
是看上上一棟跟上上上一棟哪個多(槍棟相鄰不能算)

0 1 4 8 5 16 21 22 43 28 51 76 174 131 

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

int rob(int* nums, int numsSize) { if(numsSize<=1)return nums[0]; else if(numsSize==2)return nums[0]>=nums[1]? nums[0]:nums[1]; int weight[numsSize+1]; weight[0]=0; weight[1]=nums[0]; weight[2]=nums[1]; int max=weight[1]>=weight[2]? weight[1]:weight[2]; for(int i=2;i<numsSize;i++){ weight[i+1]= weight[i+1-2]>= weight[i+1-3]? weight[i+1-2]+nums[i]:weight[i+1-3]+nums[i]; max = max>weight[i+1]? max:weight[i+1]; } //for(int i =0;i<numsSize+1;i++)printf("%d ",weight[i]); return max; }