Leetcode House Robber II

Leetcode House Robber II 相关代码，本代码使用dp算法完成相关问题，本算法的关键为把存在环的退化成不带环的情况，在不带环的情况，用动态规化是相当容易的，以下是代码以及相关测试。算法复杂度为O(n)。

#include <iostream>#include <vector>using namespace std;// the circumstance of circle can be divided into two case// First: We choose the last, which means that the nearest two of it cannot be choose. And then//        the max of it should be no_circle[1 --> i-2] + nums[i];// Second: We doesn't choose the last, which means that nearest two of it can be choose or not.//         And then, the max of it should be no_circle[0 --> i-1];// And the result of the [0 --> i] with circle is the maximum of the former two condition.class Solution {public:    int rob(vector<int>& nums) {        if (nums.size() == 0) {            return 0;        }        int len = nums.size();        vector<vector<int> > max_longer(len, vector<int>(2, 0));        max_longer[0][0] = nums[0];        int max = nums[0];        // preprocess the record before the common one        if (len <= 3) {            for (int i = 0; i < len; i ++) {                max = max > nums[i]? max : nums[i];            }        } else {            max_longer[1][0] = nums[0] > nums[1]? nums[0] : nums[1];            max_longer[1][1] = nums[1];            max_longer[2][0] = nums[0] + nums[2] > nums[1]? nums[0] + nums[2] : nums[1];            max_longer[2][1] = nums[2] > nums[1]? nums[2] : nums[1];        }        for (int i = 3; i < len; i ++) {            // for the max result            int choose = nums[i] + max_longer[i - 2][1];            int not_choose = max_longer[i - 1][0];            max = choose > max? choose : max;            max = not_choose > max? not_choose : max;           // for the from 0 to i without circle max           choose = nums[i] + max_longer[i - 2][0];           not_choose = max_longer[i - 1][0];           max_longer[i][0] = choose > not_choose? choose : not_choose;           // for the from 1 to i without circle max           choose = nums[i] + max_longer[i - 2][1];           not_choose = max_longer[i - 1][1];           max_longer[i][1] = choose > not_choose? choose : not_choose;        }        return max;    }};int main(int argc, char * argv[]) {    Solution so;    vector<int> test(5, 0);    test[0] = 1;    test[1] = 5;    int re = so.rob(test);    cout<<"max:"<<re<<endl;    return 0;}