LeetCode 198.House Robber-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_42265429/article/details/97118378

本文介绍了一种基于动态规划的算法解决方案，用于解决HouseRobber问题。该问题要求从一排房屋中挑选出最大的金钱总额，但相邻房屋不能同时被选。通过使用动态规划的方法，我们能够有效地解决这个问题。

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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.

题目大意

给定定长数组，求出选择数组中的数构成最大值，要求选择的数之间存在间隔，并且选择的数量不限

思路

动态规划的思想保存之前的状态

java version

class Solution{
	public int rob(int [] nums) {
      if(nums.length == 0) return 0;
      if(nums.length == 1) return nums[0];
      if(nums.length == 2) return Math.max(nums[0], nums[1]);
		int [] dp = new int[nums.length];
		dp[0] = nums[0];
		dp[1] = Math.max(nums[0], nums[1]);
		for(int i = 2; i < nums.length; i++) {
			dp[i] =  Math.max(dp[i - 1],  dp[i - 2]+ nums[i]);
		}
		return dp[dp.length - 1];
	}
}

C++ version

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