【LeetCode】198. House Robber

题目：

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.

题解：

　　对于求最大，最小类问题，应该联想到是否能用DP思想，问题可以具体为给定一个数组，取出一系列的数使和最大，约束就是取出的数不能相邻。状态方程：

dp[n] = max(dp[n-1], dp[n-2] + array[n]);

Solution 1

class Solution {
public:
    int rob(vector<int> &money) {
        int n = money.size();
        if (n==0) return 0;
        vector<int> dp(n, 0);
        if (n>=1) dp[0] = money[0];
        if (n>=2) dp[1] = max(money[0], money[1]);
        for (int i=2; i<n; i++){
            dp[i] = max(dp[i-1], dp[i-2] + money[i]);
        }
        return dp[n-1];
    }
};

　　可以看出Solution1中数组只需要维护一次，那么我们就可以用两个变量代表前一个最大值和前两个的最大值，

Solution 2

class Solution {
public：
    int rob(vector<int> &money) {
        int n1=0, n2=0;
        int n = money.size();
        for (int i=0; i<n; i++){
            int now = max(n1, n2 + money[i]);
            n2 = n1;
            n1 = now;
        }
        return n1;
    }
}；

　　还有一种类似比较容易理解的方法

Solution 3

class Solution {
public:
    int rob(vector<int> &money) {
        int n1 = 0, n2 = 0, n = money.size();
        for (int i=0; i<n; i++) {
            if (i%2==0) n1 = max(n1+money[i], n2);
            else n2 = max(n1, n2+money[i]);
        }
        return max(n1, n2);
    }
};