LeetCode 198 House Robber (dp)

本文介绍了一个经典的动态规划问题——打家劫舍。该问题要求在一个整数数组中找到最大的非相邻元素之和,模拟了抢劫一排房屋但不能连续抢劫相邻两家的情况。通过两种方法展示了如何有效地解决这个问题,一种使用二维数组存储中间结果,另一种通过优化空间复杂度到O(1)的方式。

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.

题目分析:简单dp,dp[i][0/1],表示第i家(不偷/偷)

public class Solution {
    public int rob(int[] nums) {
        int n = nums.length;
        if (n == 0) {
            return 0;
        }
        int[][] dp = new int[n + 1][2];
        dp[0][1] = nums[0];
        for (int i = 1; i < n; i ++) {
            dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1]);
            dp[i][1] = dp[i - 1][0] + nums[i];
        }
        return Math.max(dp[n - 1][0], dp[n - 1][1]);
    }
}

从dp递推中可以发现,本次值的决策只与上一次的状态有关,因此空间可以优化成O(1)

public class Solution {
    public int rob(int[] nums) {
        int n = nums.length;
        if (n == 0) {
            return 0;
        }
        int pre1 = nums[0], pre0 = 0, cur1 = 0, cur0 = 0;
        for (int i = 1; i < n; i ++) {
            cur0 = Math.max(pre1, pre0);
            cur1 = pre0 + nums[i];
            pre1 = cur1;
            pre0 = cur0;
        }
        return Math.max(pre0, pre1);
    }
}


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