Leetcode 740. Delete and Earn

原题链接：https://leetcode.com/problems/delete-and-earn/description/

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: 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].

Solution：

本题属于动态规划的范畴，我们需要找到最大收益数的特点，当前数所能获得的最大收益只与前一个数包含与不包含自身这两种情况分别的最大值有关，如果当前数需要拿到，那么必不能拿前一个数，也就是加上不包含前一个数的最大收益，如果当前数不需要，那么就是前一个数两种情况中的最大值作为当前不包含自身数的最大收益，具体请看代码（需要注意的是数组为空的情况）:

#include <iostream>
#include <vector>
using namespace std;

#define MaxN 10000

int max(int a, int b) {
return a > b ? a : b;
}

int deleteAndEarn(vector<int>& nums) {
if (nums.empty()) return 0;
vector<pair<int, int>> sum(MaxN, make_pair(0, 0));
int maxN = 0;
for (int i = 0; i < nums.size(); ++i) {
maxN = maxN > nums[i] ? maxN : nums[i];
sum[nums[i] - 1].first += nums[i];
}
// 动态规划的思想，当前数的最大值只与前一个包含与不包含自身的值有关
for (int i = 1; i < maxN; ++i) {
sum[i].first = sum[i - 1].second + sum[i].first;
sum[i].second = max(sum[i - 1].first, sum[i - 1].second);
}
return max(sum[maxN - 1].first, sum[maxN - 1].second);
}

int main() {
//vector<int> nums = { 2, 2, 3, 3, 3, 4 };
vector<int> nums;
cout << deleteAndEarn(nums) << endl;

system("pause");
return 0;
}