Leetcode: House Robber II

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.

因为第一个数和最后一个数不能同时选，因此需要做两次dp，第一次不选第一个数，第二次选第一个数，然后取两次结果的较大值。

代码1：

class Solution {
public:
    int rob(vector<int>& nums) {
        int i, n = nums.size();
        vector<int>table (nums.size()+1, 0);

        if (n == 0) return 0;
        if (n == 1) return nums[0];

        //第一遍
        table[2] = nums[1];
        for (i = 2; i < n; ++i) {
            table[i+1] = max(table[i], table[i-1]+nums[i]);
        }

        //第二遍
        table[1] = nums[0];
        for (i = 1; i < n-1; ++i) {
            table[i+1] = max(table[i], table[i-1]+nums[i]);
        }

        return max(table[n-1], table[n]);
    }
};

代码2：

class Solution {
public:
    int rob(vector<int>& nums) {
        int i, pre, cur, tmp, result, n = nums.size();

        if (n == 0) return 0;
        if (n == 1) return nums[0];

        //第一遍
        pre = 0;
        cur = nums[1];
        for (i = 2; i < n; ++i) {
            tmp = cur;
            cur = max(pre+nums[i], cur);
            pre = tmp;
        }
        result = cur;

        //第二遍
        pre = 0;
        cur = nums[0];
        for (i = 1; i < n-1; ++i) {
            tmp = cur;
            cur = max(pre+nums[i], cur);
            pre = tmp;
        }

        return max(result, cur);
    }
};