198.House Robber - Leetcode题解

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20.Valid Parentheses

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198.House Robber

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414.Third Maximum Number

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441.Arranging Coins

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572.Subtree of Another Tree

575.Distribute Candies

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627.Swap Salary

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643.Maximum Average Subarray I

645.Set Mismatch

653.Two Sum IV - Input is a BST

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661.Image Smoother

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669.Trim a Binary Search Tree

671.Second Minimum Node in a Binary Tree

680.Valid Palindrome II

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687.Longest Univalue Path

690.Employee Importance

693.Binary Number with Alternating Bits

696.Count Binary Substring

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783.Minimum Distance Between BST Nodes

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829.Consecutive Numbers Sum

830.Position of Large Groups

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994.Rotting Oranges

997.Find the Town Judge

999.Available Captures for Rook

1002.Find Common Characters

1005.Maximise Sum of Array after K Negations

1008.Construct Binary Search Tree from Preorder Traversal

1021.Remove Outermost Parentheses

1025.Divisor Game

1029.Tow City Scheduling

1030.Matrix Cells in Distance Order

1033.Moving Stones Until Consecutive

1037.Valid Boomerang

1038.Binary Search Tree to Greater Sum Tree

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1089.Duplicate Zeros

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1018.Binary Prefix Divisible By 5

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1128.Number of Equivalent Domino Pairs

1169.Invalid Transactions

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1282.用户分组

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1379.Find a Corresponding Node of a Binary Tree in a Clone of That Tree

1395.Count Number of Teams

面试题 16.01

面试题 16.18

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198.House Robber

198.House Robber
难度:Easy
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.
要至少间隔1个，其实也可以看出，至多间隔两个。
简单暴力的方法，直接递归，四个一循环。
class Solution {
  int rob(vector<int>& nums, int start) {
        if(nums.empty()) return 0;
        switch(nums.size() - start)
        {
            case 1: return nums[start];
            case 2: return max(nums[start],nums[start+1]);
            case 3: return max(nums[start]+nums[start+2],nums[start+1]);
            case 4: return max(max(nums[start]+nums[start+2], nums[start+1]+nums[start+3]), nums[start]+nums[start+3]);
            default: break;
        }
      return max(nums[start]+rob(nums,start+2), nums[start+1]+rob(nums, start+3));
  }
    
public:
    int rob(vector<int>& nums) {
        return rob(nums,0);        
        
    }
};
不过会显示超时。
另一种方法，采用动态规划的方法，第n项的和可以看做是前n-2项加上第n项，与前n-1项和的较大值。
class Solution {
public:
    int rob(vector<int>& nums) {
        if(nums.empty()) return 0;
        int prenum=0;
        int curnum=0;
        for(auto i:nums)
        {
           int tmp= prenum+i;
            prenum=curnum;
            curnum=max(tmp,curnum);            
            
        }
                     
        return max(curnum,prenum);
    }
};
执行用时 :4 ms, 在所有 C++ 提交中击败了80.13%的用户
内存消耗 :8.4 MB, 在所有 C++ 提交中击败了93.46%的用户

197.Rising Temperature

202.Happy Number

Last modified 7mo ago

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