LeetCode 198. House Robber 打家劫舍

198. House Robber

题目描述

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.

示例:

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.

解答

这道题是经典的DP问题，对于一个房屋，小偷可以选择偷或者不偷，如果偷了当前房屋，小偷就不能偷下一个房屋。

假设小偷站在一个房屋前面，他当前可以偷钱的最大值等于max(他站在上一个房屋前面可以偷的最大值， 他偷当前房屋的钱 + 他站在上上一个房屋前面可以偷的最大值)。

代码

class Solution {
public:
    int rob(vector<int>& nums) {
        if(nums.size() == NULL) return NULL;
  		auto _size = nums.size();
  		memo = vector<int>(_size + 1, -1);
  		memo[0] = 0;
  		memo[1] = nums[0];
  		for (int i = 2; i <= _size; i++) {
  			memo[i] = max(memo[i-1], memo[i - 2] + nums[i-1]);
  		}
  		return memo[_size];


    }
private:
	vector<int> memo;    

}

因为发现memo数组只需要记录memo[i - 1] 和memo[i - 2] 的值就行了，可以把这两个值用 pre 和 now 代替，简化代码如下。

class Solution {
public:
    int rob(vector<int>& nums) {
    	if(nums.size() == NULL) return NULL;
  		auto _size = nums.size();
  		
  		auto pre = 0;
  		auto now = nums[0];
  		for (int i = 2; i <= _size; i++) {
  			auto temp = now;
  			now = max(now, pre + nums[i - 1]);
  			pre = temp;
  		}
  		return now;


    }
private:
	vector<int> memo;    
};