[LeetCode] House Robber

发布时间：2020-12-13 20:45:14 所属栏目：PHP教程来源：网络整理

导读：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

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.

解题思路

动态计划
状态转移方程：f[i] =max(f[i⑴],f[i⑵]+c[i])
f[i]表示进入第i+1个房间时所得到的最大财富。
为了节省空间，只是用3个变量prepre,pre,cur便可。

实现代码1

//Runtime:2ms
class Solution {
public:
    int rob(vector<int>& nums) {
        int len = nums.size();
        if (len == 0) return 0;
        if (len == 1) return nums[0];
        if (len == 2) return max(nums[0],nums[1]);
        int prepre = nums[0];
        int pre = max(nums[0],nums[1]);
        int cur;
        for (int i = 2; i < len; i++)
        {
            cur = max(pre,prepre + nums[i]);
            prepre = pre;
            pre = cur;
        }
        return cur;
    }
};

实现代码2

# Runtime:76ms
class Solution:
    # @param {integer[]} nums
    # @return {integer}
    def rob(self,nums):
        size = len(nums)
        if size == 0:
            return 0
        elif size == 1:
            return nums[0]
        elif size == 2:
            return max(nums[0],nums[1])

        prepre = nums[0]
        pre = max(nums[0],nums[1])
        for i in range(2,size):
            cur = max(pre,prepre + nums[i])
            prepre = pre
            pre = cur

        return cur

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!