Leetcode-198. House Robber

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

——————————————————————————————

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.
说实话，这个题还是有点蛋疼的，一开始我打算用暴力解，复杂度是O(2^n)。后来改进了一下O(n)的时间复杂度，空间复杂度O(1)
public class Solution {
public int rob(int[] nums) {
int currentMax = 0, value_1 = 0, value_2 = 0;
for(int i = 0 ; i < nums.length ;i ++){

int temp = value_1;
value_1 = value_2;
if(temp == value_2){
value_2 = temp + nums[i];
}else{
value_2 = temp + nums[i];
}
value_2 = Math.max(value_1,value_2);

}
currentMax = Math.max(value_1,value_2);

return currentMax;
}
}