House Robber

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.

class Solution {
public:
    int rob(vector<int>& nums) {

    }
};

理解

题目大致可以理解为，给出数组nums，我们要在其中找出和最大的两两元素不相邻的子集；
第一想到的是之前解和最大子集的方式，从nums[0]开始遍历，每次选出最大的子集即可，时间复杂度为O ( n ) ;
代码如下：

class Solution {
public:
    int rob(vector<int>& nums) {
       int a = 0;
        int b = 0;

        for (int i=0; i<nums.size(); i++)
        {
            if (i%2==0)
            {
                a = max(a+nums[i], b);
            }
            else
            {
                b = max(a, b+nums[i]);
            }
        }

        return max(a, b); 
    }
};

改进

显然看了detail后：

参考了 0 ms 的代码：

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 0)
            return 0;
        if (nums.size() == 1)
            return nums[0];
        if (nums.size() == 2)
            return max(nums[0], nums[1]);

        int result0 = nums[0];
        int result1 = max(nums[0], nums[1]);
        int temp = 0;

        for (int i = 2; i <nums.size(); i++)
        {
            temp = max(result0+nums[i], result1);
            result0 = result1;
            result1 = temp;
        }
        return result1;
    }
};

其实仅仅是对于nums.size( ) = 0 , 1 的改进，就大大缩短了延迟；