LeetCode-House Robber-解题报告

dinglin0xff

于 2015-07-02 14:21:37 发布

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分类专栏： leetcode 文章标签： C++ leetcode 动态规划

本文链接：https://blog.youkuaiyun.com/dd523762588/article/details/46726139

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本文针对LeetCode上的House Robber问题进行了解析，通过动态规划的方法解决了如何在不触动相邻房屋警报的情况下，获取最大金额的问题。介绍了状态转移方程，并给出了具体的C++实现代码。

原题链接https://leetcode.com/problems/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.

简单dp

转移方程 dp[i] = max(dp[i - 2] + nums[i], dp[i - 1]);

dp[i]表示走到第i房子所取得的最大值。

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

		return dp[nums.size() - 1];
		
	}	
};