zoukankan      html  css  js  c++  java
  • House Robber

    题目描述:

    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.

    这明显是个动态规划问题,直接上代码。

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

        由于声明了一个数组,所以空间复杂度为O(n)。当然,我们可以把空间复杂度降到O(1),因为每次我们只需要数组最新的两个元素。不过,这道题还有一种解题思路:对于每个房子,我们有两个选择,抢或不抢。如果抢了,说明前面的房子我们没抢;如果没抢,说明前面的房子我们可能抢了也可能没抢,此时选择两种情况下钱多的那种情况。

    int rob(vector<int>& nums) {
        int n = nums.size();
        if (n == 0)
            return 0;
        if (n == 1)
            return nums[0];
        if (n == 2)
            return max(nums[0], nums[1]);
    
        int a = nums[0];    // rob current
        int b = 0;          // not rob current
        for (int i = 1; i < n; ++i)
        {
            int tmp = max(a, b);
            a = b + nums[i];
            b = tmp;
        }
        return max(a, b);
    }

    可参见 https://leetcode.com/discuss/30018/o-n-dynamic-solution-with-o-1-memory

        此问题还有一个变种:House Robber II ,这种情况下,n个房子是连成环的。直接给答案了:

    int rob(vector<int>& nums) {
        int n = nums.size();
        if (n == 0)
            return 0;
        if (n == 1)
            return nums[0];
        if (n == 2)
            return max(nums[0], nums[1]);
        // rob 1 to n-1
        int a = nums[0];
        int b = 0;
        for (int i = 1; i < n-1; ++i)
        {
            int tmp = max(a, b);
            a = b + nums[i];
            b = tmp;
        }
        int first = max(a, b);
        // rob 2 to n
        a = nums[1];
        b = 0;
        for (int i = 2; i < n; ++i)
        {
            int tmp = max(a, b);
            a = b + nums[i];
            b = tmp;
        }
        int second = max(a, b);
    
        return max(first, second);
        
    }
  • 相关阅读:
    面试可能遇到的关联式容器(map、set等)相关问题
    C++学习 STL组件之vector部分总结
    C++ 菱形虚拟继承 与 指针偏移问题
    C++ 关键字 explicit 的使用
    C++ sort()排序函数用法
    C++ 字符流 stringstream
    C/C++ 每日一题
    C/C++ 超长正整数相加
    C++ 二叉搜索树原理及其实现
    Tomcat安装
  • 原文地址:https://www.cnblogs.com/gattaca/p/4755131.html
Copyright © 2011-2022 走看看