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.

Example 1:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

题意：一个数组隔项求和，求结果的最大值

分析：

一、隔项求和，简单想，直接判断奇偶(只要不是相邻项)再求和

二、DP,转移状态公式：dp[i]=max(dp[i-1], dp[i-2]+nums[i]), i>=2;

                                      dp[0]=nums[0],dp[1]=max(nums[0], nums[1]);

/*
* @Author: yufeng
* @Date:   2018-10-13 11:24:23
* @Last Modified by:   yufeng
* @Last Modified time: 2018-10-13 11:33:29
* @Email: [e-mail]
* @Blog: http://fzhiy.com/
*/
class Solution {
public:
    int rob(vector<int>& nums) {
        int suma=0,sumb=0;
        for(int i=0; i<nums.size(); i++){
            if(i&1){
                suma=max(sumb, suma+nums[i]);
            }else {
                sumb=max(suma, sumb+nums[i]);
            }
        }
        return max(suma, sumb);
    }
};

 

//DP
class Solution {
public:
    int rob(vector<int>& nums) {
        int len = nums.size();
        if(!len) return 0;
        if(len==1) return nums[0];
        if(len==2) return max(nums[0], nums[1]);
        std::vector<int> dp(len, 0);
        dp[0]=nums[0],dp[1]=max(nums[0], nums[1]);
        for(int i=2; i<len; i++){
            // 要么前一项dp[i-1]大，要么是前前一项dp[i-2]+nums[i]大
            dp[i]=max(dp[i-1], dp[i-2]+nums[i]);
        }
        return dp[len-1];
    }
};