740. Delete and Earn

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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].

题目大意

给定一个序列nums，从中删除所有nums[i]并加入总分，同时舍弃所有nums[i+1]和nums[i-1]。求出总分最高的选择。

解题思路

首先，需要对序列进行从小到大的排序。
然后，使用动态规划的思路。
在遍历以排序数组时，我们考虑以下三种情况:
1. 当前值与前项值相等
2. 当前值与前项值相差为1
3. 当前值与前项值相差大于1
将序列分为两个部分:
1. preciousPart: 在选取当前值时，所得到的最大值。
2. lastPart: 在不选取当前值时，所得到的最大值。

所以，这个算法步骤为:
1. 当前值与前项值相等，preciousPart加上当前值nums[i]。
2. 当前值与前项值相差为1，修改preciousPart，结果为lastPart加上当前值nums[i]，同时，lastPart修改为不选取当前值时，所得到的最大值。
3. 当前值与前项值相差大于1，修改preciousPart，结果为当前的最大值maxCache加上当前值nums[i]，同时lastPart修改为不选取当前值时，所得到的最大值。

以此类推，当遍历到最后一个数字时，比较选取当前值时所得到的最大值preciousPart和不选取当前值时所得到的最大值lastPart。

算法复杂度

O(N)

代码实现

class Solution {public:    int deleteAndEarn(vector<int>& nums) {        int length = nums.size();        if (length == 0) return 0;        sort(nums.begin(), nums.end());        int preciousPart = 0, lastPart = 0, preciousNum = 0;        for (int i = 0; i < length; i++) {            int maxCache = max(preciousPart , lastPart);            if (i > 0 && nums[i] == nums[i-1]) {                preciousPart += nums[i];            } else if (preciousNum+1 == nums[i]) {                preciousPart = lastPart + nums[i];                lastPart = maxCache;            } else {                preciousPart = maxCache + nums[i];                lastPart = maxCache;            }            preciousNum = nums[i];        }        return max(preciousPart, lastPart);    }};