LeetCode 198: House Robber

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.

题目本质：求一个数组中元素的最大和，但是元素不能从连续的位置取。可用动态规划的思想解决：保存在每个位置时，加上当前元素和不加当前元素能得到的最大元素和。

public class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0)
            return 0;
        int[][] dp = new int[nums.length][2];
        //0表示不被抢劫，1表示被抢劫
        for(int i = 0; i < dp.length; i++){
            dp[i][0] = 0;
            dp[i][1] = 0;
        }
   
        dp[0][0] = 0;
        dp[0][1] = nums[0];
        
        for(int i = 1; i < dp.length; i++){
            //如果不加入，保持抢劫到前一家时的最大和；如果加入，上一家必须没有被抢劫过
            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[dp.length-1][0], dp[dp.length-1][1]);
    }
}

还有更简单的方法：用temp0表示不抢劫此家的最大和，temp1表示抢劫此家的最大和。

public class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0)
            return 0;

        int temp0 = 0, temp1 = nums[0];
        int temp = 0;

        for(int i = 1; i < nums.length; i++){
            temp = Math.max(temp0, temp1);
            temp1 = temp0 + nums[i];
            temp0 = temp;
        }

        return Math.max(temp0, temp1);
    }
}