dp/house-robber

House Robber

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.

分析

这是一个多阶段最优化问题，且要走到最底部才能知道答案，因此广搜排除，只剩下贪心和动规。贪心明显要排除，只剩下动规。

设状态 f[i] 为到位置i时能抢到的金钱最大和，那么状态转移方程如下：

f[i]=max(f[i-1], f[i-2] + nums[i])

其含义是，如果不选择i，则抢到的钱是f[i-1]，如果选择i，则能抢到的钱是f[i-2] + nums[i]。

解法 1

// House Robber
// Time Complexity: O(n), Space Complexity: O(n)
public class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];

        int[] f = new int[nums.length];
        f[0] = nums[0];
        f[1] = Math.max(nums[0], nums[1]);

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

解法 2

在状态转移方程中，我们可以发现 f[i]仅仅依赖前两项，因此用两个整数变量即可代替一位数组，将空间复杂度降为O(1)。

// House Robber
// Time Complexity: O(n), Space Complexity: O(1)
public class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];

        int even = nums[0];
        int odd = Math.max(nums[0], nums[1]);

        for (int i = 2; i < nums.length; ++i) {
            if (i % 2 == 0) {
                even = Math.max(even + nums[i], odd);
            } else {
                odd = Math.max(odd + nums[i], even);
            }
        }
        return Math.max(even, odd);
    }
}