House Robber

1 House Robber

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

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.

public class Solution {    public int rob(int[] nums) {        int n = nums.length;        // 没有房子的时候收入是0        if (n == 0)            return 0;        // 只有一个房子的时候选择唯一        else if (n == 1) {            return nums[0];        } else {            int[] maxRob = new int[n];            maxRob[0] = nums[0];            // 当有两个房子的时候，只能二选一，当然选钱多的那一个            maxRob[1] = Math.max(nums[0], nums[1]);            // 房子从0开始编号            // 当考虑是否抢劫第i号房子的时候，有两种情况：            // （1）抢i，则i-1不能抢，最大收益是 nums[i] + maxRob[i-2]            // （2）不抢i，则最大收益是maxRob[i-1]            for (int i = 2; i < n; i++) {                maxRob[i] = Math.max(nums[i] + maxRob[i - 2], maxRob[i - 1]);            }            return maxRob[n - 1];        }    }}

---

---

2 House Robber II

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.

public class Solution {    public int rob(int[] nums) {        int n = nums.length;        if (n == 0)            return 0;        else if (n == 1) {            return nums[0];        } else {            //如果抢最后一个房子，则第一个房子不能抢；            //最大收益为，rob1中，抢1到n-1号房子的最大收益            int[] nums1 = Arrays.copyOfRange(nums, 1, n);            int maxNoFirst = rob1(nums1);            //如果不抢最后一个房子，则第一个房子可以抢；            //最大收益为，rob1中，抢0到n-2号房子的最大收益            int[] nums2 = Arrays.copyOfRange(nums, 0, n-1);            int maxNoLast = rob1(nums2);            return Math.max(maxNoFirst, maxNoLast);        }    }    public int rob1(int[] nums) {        int n = nums.length;        // 没有房子的时候收入是0        if (n == 0)            return 0;        // 只有一个房子的时候选择唯一        else if (n == 1) {            return nums[0];        } else {            int[] maxRob = new int[n];            maxRob[0] = nums[0];            // 当有两个房子的时候，只能二选一，当然选钱多的那一个            maxRob[1] = Math.max(nums[0], nums[1]);            // 房子从0开始编号            // 当考虑是否抢劫第i号房子的时候，有两种情况：            // （1）抢i，则i-1不能抢，最大收益是 nums[i] + maxRob[i-2]            // （2）不抢i，则最大收益是maxRob[i-1]            for (int i = 2; i < n; i++) {                maxRob[i] = Math.max(nums[i] + maxRob[i - 2], maxRob[i - 1]);            }            return maxRob[n - 1];        }    }}