Delete and Earn

问题来源

问题描述

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: 6Explanation: 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: 9Explanation: 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].

问题分析

根据题意，我们首先知道，当我们选择了nums[i]， 那么我们就需要抛弃 nums[i]-1与nums[i]+1，换而言之，我们就需要抛弃其相邻元素。那么如何取舍就是此题的重点。显然，如果我们选择了一个数字，那么我们实际上也就需要将这个数所能带来的价值计算出来。例如Example2里，我们选择了3，那么我们就需要计算出3能带来的价值，即 3+3+3 = 9. 因此，我们首先生成一个数组val，令 val[nums[i]] += nums[i],接下来的问题就变成了，如何取val中的元素使得和最大（不能连续取值）。
我们令 dp[i] 代表取到val中第i个元素，则有：
dp[i] = dp[i-1]，如果取了val[i-1]
dp[i] = dp[i-2]+val[i]， 如果没取val[i-1]
那么按照我们的意愿，当然是
dp[i] = max(dp[i-1],dp[i-2]+val[i])
问题到此解决

解决代码

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