Leetcode 198. House Robber (Easy) (cpp)

最新推荐文章于 2024-01-18 14:02:27 发布

Niko_Ke

最新推荐文章于 2024-01-18 14:02:27 发布

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本文介绍了解决LeetCode 198题“打家劫舍”的方法，这是一个经典的动态规划问题。文章详细解释了如何通过动态规划算法来计算在不连续抢劫相邻房屋的情况下能获得的最大金额。

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Leetcode 198. House Robber (Easy) (cpp)

Tag: Dynamic Programming

Difficulty: Easy

/*

198. House Robber (Easy)

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) {
    	int n = nums.size();
    	switch(n) {
    		case 0: return 0; break;
    		case 1: return nums[0]; break; 
    	}
        vector<int> res(n, 0);
        res[0] = nums[0];
        res[1] = max(nums[0], nums[1]);
        for (int i = 2; i < n; i++) {
        	res[i] = max(nums[i] + res[i - 2], res[i - 1]);
        }
        return res[n - 1];
    }
};