740. Delete and Earn

LeetCode 740: 删除并获得点数的DP解法
本文介绍了解决LeetCode第740题“删除并获得点数”的动态规划算法。通过将问题转化为经典的House Robber问题,我们能够有效地求解出最大点数。文章提供了详细的解析过程和C++代码实现。

Given an array nums of integers, you can perform operations on the array.

In each operation, you pick any nums[i] and delete it to earn nums[i] points. After, you must delete every element equal to nums[i] - 1 or nums[i] + 1.

You start with 0 points. Return the maximum number of points you can earn by applying such operations.

 

Example 1:

Input: nums = [3, 4, 2]
Output: 6
Explanation: 
Delete 4 to earn 4 points, consequently 3 is also deleted.
Then, delete 2 to earn 2 points. 6 total points are earned.

 

Example 2:

Input: nums = [2, 2, 3, 3, 3, 4]
Output: 9
Explanation: 
Delete 3 to earn 3 points, deleting both 2's and the 4.
Then, delete 3 again to earn 3 points, and 3 again to earn 3 points.
9 total points are earned.

 

Note:

  • The length of nums is at most 20000.
  • Each element nums[i] is an integer in the range [1, 10000].

 

Approach #1: DP. [C++]

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        int len = nums.size();
        if (len == 0) return 0;
        int r = *max_element(nums.begin(), nums.end());
        vector<int> points(r+1, 0);
        for (int num : nums) 
            points[num] += num;
        return solve(points);
    }
    
private:
    int solve(const vector<int>& points) {
        int dp1 = 0, dp2 = 0;
        for (int point : points) {
            int dp = max(dp2 + point, dp1);
            dp2 = dp1;
            dp1 = dp;
        }
        
        return dp1;
    }
};

  

Analysis:

If we take nums[i], we can safely take all of its copies. We can't take any of copies of nums[i-1] and nums[i+1], This problem is reduced to 198 House Robber.

 

Houses[i] has all the copies of num whose value is i.

[3, 4, 2] -> [0, 2, 3, 4], rob([0, 2, 3, 4]) = 6

[2, 2, 3, 3, 3, 4] -> [0, 2*2, 3*3, 4], rob([0, 2*2, 3*3, 4]) = 9

 

Time complexity: O(n+r) reduction + O(r) solving rob = O(n + r)

Space complexity: O(r).

 

Reverence:

http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-740-delete-and-earn/

 

转载于:https://www.cnblogs.com/ruruozhenhao/p/10533986.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值