LeetCode : 198. House Robber 入室抢劫-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_16234613/article/details/89888906

博客围绕房屋抢劫问题展开，给定代表每间房屋藏钱数的非负整数列表，需确定不触发警报情况下可抢劫的最大金额。通过动态规划解决，考虑抢劫和不抢劫当前位置两种情况，并给出示例及对应输出。

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试题
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.

Example 1:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.
Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.
代码
以当前的位置最大可以获得的钱作为动归。存在两种情况：抢劫当前位置和不抢劫当前位置。

class Solution {
    public int rob(int[] nums) {
        if(nums==null || nums.length==0) return 0;
        
        int[] count = new int[nums.length];
        for(int i=0; i<nums.length; i++){
            count[i] = nums[i];
            // for(int j=0; j<i-1; j++){
            //     count[i] = Math.max(count[i], count[j]+nums[i]);
            // }
            if(i-2>=0)
                count[i] = Math.max(count[i], count[i-2]+nums[i]);
            if(i-1>=0)
                count[i] = Math.max(count[i], count[i-1]);
        }
        return count[nums.length-1];
    }
}