题目描述

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.

Credits:
Special thanks to @ifanchu for adding this problem and creating all test cases. Also thanks to @ts for adding additional test cases.

分析

本质：求数组不相邻元素最大和

动态规划（背包问题）：设P[i]表示从0~i个房间抢劫的最大收益。

P[i]={nums[i]+P[i−2]，偷第i个P[i−1]，不偷第i个

每次迭代只需要P的两个元素，并不需要设数组P。设两个变量为：

take :nums[i] + P[i-2]
nonTake:P[i-1]

代码

时间复杂度是O(n)，空间复杂度是O(n)的代码：

    public static int rob2(int[] nums) {

        if (nums.length == 0) {
            return 0;
        }

        if (nums.length == 1) {
            return nums[0];
        }

        int[] P = new int[nums.length];

        P[0] = nums[0];
        P[1] = Math.max(nums[0], nums[1]);

        for (int i = 2; i < nums.length; i++) {
            P[i] = Math.max(nums[i] + P[i - 2], P[i - 1]);
        }

        return P[nums.length - 1];
    }

时间复杂度是O(n)，空间复杂度是O(1)的代码：

    public static int rob(int[] nums) {

        int take = 0;
        int nonTake = 0;
        int max = 0;

        for (int i = 0; i < nums.length; i++) {
            take = nums[i] + nonTake;
            nonTake = max;
            max = Math.max(take, nonTake);
        }

        return max;
    }