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.
分析：这题是上一题House Robber的升级版～～新加了环形的街道biu biu biu～～
所以就会考虑到，最后一个和第一个房子是不能够同时进入的~要不然会告诉警察叔叔～～
所以分为两种情况～～
0.不包括最后一个屋子～就抢劫0~n-2号屋子～
1.不包括第一个屋子～就抢劫1~n-1号屋子～
这样的话，return上面两种情况的最大值就好了～调用两次子函数求值，主函数取其最大值返回～
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class Solution { public: int rob(vector<int>& nums) { int n = nums.size(); if(n == 0) return 0; if(n == 1) return nums[0]; if(n == 2) return max(nums[0], nums[1]); return max(func(nums, 0, n-2), func(nums, 1, n-1)); } int func(vector<int>& nums, int begin, int end) { int n = end - begin + 1; vector<int> dp(n); dp[0] = nums[begin]; dp[1] = max(nums[begin], nums[begin+1]); for(int i = 2; i < n; i++) { int temp = dp[i - 2] + nums[begin+i]; dp[i] = max(temp, dp[i-1]); } return dp[n - 1]; } }; |
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