House Robber II

题目

原题

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.

思路

这道题是House Robber(链接:https://leetcode.com/problems/house-robber/)的升级版本.即所给定的nums构成了环.由于每个house最多被rob一次, 设len=nums.size(),所以这个环只对nums[0]和nums[len-1]起作用.即如果选择了nums[0], 就不能选择了nums[len-1];如果选择了nums[len-1], 就不能选择nums[0].所以该问题就变了了House Robber的两个子问题的结果最大值.ans=max(rob(nums[0,...,len-2), rob(nums[1,...,len-1)). 其中rob为House Robber问题的解.具体看代码.

code

class Solution {public:    int rob_helper(const vector<int>& nums, const int s, const int e) {        int len = e - s + 1;        vector<int>res(len, 0);        res[0] = nums[s];        res[1] = max(nums[s], nums[s + 1]);        for(int i = s + 2; i <= e; i++) {            res[i - s] = max(res[i - s - 2] + nums[i], res[i - s - 1]);        }        return res[len - 1];    }    int rob(vector<int>& nums) {        int len = nums.size();        if(len <= 0) return 0;        if(len <= 1) return nums[0];        // Include the first one of nums without the last one.        int f = rob_helper(nums, 0, len - 2);        // Include the last one of nums without the first        int s = rob_helper(nums, 1, len - 1);        return max(f, s);    }};