LintCode 534 · House Robber II (环形动态规划好题)

534 · House Robber II
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Description
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. Each house holds a certain amount of money. The only constraint you face is that adjacent houses are equipped with interconnected anti-theft systems, which will automatically alarm when two adjacent houses are robbed on the same day.

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.

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This is an extension of House Robber.It just turns a straight line into a circle.

Example
Example1

Input: nums = [3,6,4]
Output: 6
Example2

Input: nums = [2,3,2,3]
Output: 6

解法1：用DP。分house[0]被抢和不被抢两种情况。dp[n]应该还可以空间优化，也就是说只用3个变量prepre, pre, cur就可以。

class Solution {
public:
    /**
     * @param nums: An array of non-negative integers.
     * @return: The maximum amount of money you can rob tonight
     */
    int houseRobber2(vector<int> &nums) {
        int n = nums.size();
        if (n == 0) return 0;
        if (n == 1) return nums[0];
        int dp[n] = {0}; //dp[i]: the maximum money can be robbed for the first i houses.
        
        //the 0th house is not robbed
        dp[1] = nums[1];
        for (int i = 2; i < n; i++) {
            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
        }
        int res = dp[n - 1];
        //the 0th house is robbed
        dp[0] = nums[0];
        dp[1] = nums[0];  //注意这一行！
        for (int i = 2; i < n - 1; i++) { //0th house is robbed, so i < n - 1
            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);
        }
        return max(res, dp[n - 2]);
    }
};