740. Delete and Earn

740. Delete and Earn

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);
}
};