LeetCode【198】 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.

int rob(vector<int> &num) {
    if(num.size()==0)
        return 0;
    if(num.size()==1)
        return num[0];
    cout<<num.size()<<endl;
    vector<int> total(num.size(),0);
    total[0]=num[0];
    total[0]=num[0];
    if(num[1]>num[0])
        total[1]=num[1];
    else
        total[1]=num[0];
    int t=0;
    for(int i=2;i<num.size();i++)
    {
        t=num[i]+total[i-2];
        if(t>total[i-1])
            total[i]=t;
        else
            total[i]=total[i-1];
    }
    int max=0;
    for(i=total.size()-1;i>=0;i--)
    {
        if(total[i]>max)
            max = total[i];
    }
    return max;
}

然后看了些别人写的，羞愧啊。

【转】一个显然的dp。 best0表示必须不选择下一个房间，best1表示可以选择下一个房间的最大收益。
best1' = best0 因为不选第i个 在下一个房间就可以选了
best0' = max(best0, best1 + num[i]) 尝试选择num[i]，如果不更优，还不如不选它…… 

 int rob(vector<int> &num) {
        int best0 = 0, best1 = 0;
        for (int i = 0; i < num.size(); ++i) {
            int temp = best1 + num[i];
            best1 = best0;
            best0 = max(best0, temp);
        }
        return max(best0, best1);