213. House Robber II

Note: This is an extension of House Robber.

After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.

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.

这个题目相比上一个198. House Robber ,把数组变成了环形的,也就是首尾不能同时取到,那么可以分两种情况来考虑。

class Solution {
public:
    int rob(vector<int>& nums) {
        int n=nums.size();
        if(n==0)return 0;
        else if(n==1)return nums[0];
        else if(n==2) return max(nums[0],nums[1]);
        else return max(rob_find(nums,0,n-1),rob_find(nums,1,n));
    }
     int rob_find(vector<int>& nums,int l,int r) {
        int n = nums.size();
        int *a = new int[n]; 
        int *b  = new int[n]; 
        a[l] = nums[l];
        b[l] = 0;
        for(int i=l+1; i<r; i++) {
            a[i] = b[i-1] + nums[i];
            b[i] = max(a[i-1], b[i-1]);
        }
        return max(a[r-1], b[r-1]);
    }
};

你可能感兴趣的:(213. House Robber II)