LeetCode House Robber

最新推荐文章于 2021-05-17 11:01:03 发布

LarryNLPIR

最新推荐文章于 2021-05-17 11:01:03 发布

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分类专栏： ACM-动态规划 LeetCode

本文链接：https://blog.youkuaiyun.com/yangliuy/article/details/45148549

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本文探讨了一个经典的动态规划问题——打家劫舍问题。问题设定为抢劫者不能连续抢劫相邻房屋以避免触发警报。文章通过动态规划的方法提供了解决方案，并给出了两种实现方式：一种使用数组存储中间结果，另一种则采用更高效的O(1)空间复杂度的解决方案。

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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.

思路分析：典型的动态规划，和求最大和子数组有点类似，global[i] 表示num[0...i]之间的最优解，那么DP方程可以写作global[i] = Max(global[i-2] + num[i], global[i-1]),分别对应于取num[i]（此时不能取num[i-1]）和不取num[i]的最优解，然后取max即可。是一个一维DP，时间和空间复杂度都是O（n）。

AC Code

public class Solution {
    public int rob(int[] num) {
        //1218
        int n = num.length;
        if(n == 0) return 0;
        if(n == 1) return num[0];
        
        int []global = new int [n];
        global[0] = num[0];
        global[1] = Math.max(num[0], num[1]);
        for(int i = 2; i < n; i++){
            global[i] = Math.max(global[i-2] + num[i], global[i-1]);//Max(chooseNum[i], notChooseNum[i])
        }
        return global[n-1];
        //1224
    }
}

优化后只有O（1）空间的代码如下,分别用a和b对奇数index和偶数index的数取local max，每次计算要更新全局max。

 public int rob(int[] num) {
        if(num == null || num.length  == 0) return 0;
        int a = 0; //odd
        int b = 0; //even
        
        for(int i = 0; i < num.length; i++){
            if(i%2 == 0){
                b += num[i];
                b = Math.max(a, b);
            } else{
                a += num[i];
                a = Math.max(a, b);
            }
        }
        return Math.max(a,b);
    }