zoukankan      html  css  js  c++  java
  • 198. House Robber(动态规划)

    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.

    分析
    首先边界是dp[0]=nums[0],dp[1]=max(nums[0],nums[1]);对每一个house来说,只有抢与不抢两种情况,故动态转移方程dp[n]=max(dp[n-1],dp[n-2]+nums[n]).

    class Solution {
    public:
        int rob(vector<int>& nums) {
            if(nums.size()==0) return 0;
            else if(nums.size()==1) return nums[0];
            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];
        }
    };
    

    空间复杂度为O(1)的

    class Solution {
    public:
        int rob(vector<int>& nums) {
            int a=0,b=0;
            for(int i=0;i<nums.size();i++){
                 if(i%2==0)
                    a=max(a+nums[i],b);
                else 
                    b=max(b+nums[i],a);
            }
            return max(a,b);
        }
    };
    
  • 相关阅读:
    day10作业
    day9 函数作业
    Python编码及文件练习题
    day10函数命名空间,嵌套,闭包
    Python基础数据类型考试题
    day9 函数
    day8 文件操作
    day7 集合
    day6 编码
    day5 作业自我完成版
  • 原文地址:https://www.cnblogs.com/A-Little-Nut/p/8323456.html
Copyright © 2011-2022 走看看