198.House Robber （求非相邻正整数的和最大值）

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本文详细介绍了专业抢劫者在规划抢劫整条街道上的房子时，如何避免相邻房屋触发报警系统，确保行动成功。通过使用一种巧妙的动态规划算法，抢劫者能够在不被发现的情况下最大化抢劫金额。文中提供了具体步骤和代码实现，帮助读者理解并应用这一策略。

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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：新建一个数组temp[i]，长度为n-1; temp[i]表示第i个数之前的最大值，取temp[i-1]与nums[i]+temp[i-2]的较大值

temp[i] = Math.max(temp[i-1],nums[i]+temp[i-2]);

代码：

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

result = temp[nums.length-1];
}
return result;
}
}