[leetcode] 740. Delete and Earn

Description

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].

分析

题目的意思是：给你一个数组，每删除一个数，你就会获得那个位置的数值，但是同时你要删一个那个数-1的数，或者该数值+1的数，然后求你最终能够获得的最大的值。

• 我们首先统计每个数的求和，然后就可以转化成house robber的问题了。dp[i]在于拿或者不拿当前的数得到的最大的数值。
• 这道题我做不出来，看来能够正确理解题意，对解决问题的帮助很大。

代码

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        vector<int> sums(10001,0);
        for(auto num:nums){
            sums[num]+=num;
        }
        vector<int> dp(sums.size(),0);
        dp[1]=sums[1];
        for(int i=2;i<sums.size();i++){
            dp[i]=max(dp[i-1],dp[i-2]+sums[i]);
        }
        return dp[10000];
    }  
};

参考文献

Delete and Earn问题及解法
[LeetCode] Delete and Earn 删除与赚取