740 Delete and Earn

740. Delete and Earn

1. Question

Given an arraynumsof integers, you can perform operations on the array.

In each operation, you pick anynums[i]and delete it to earnnums[i]points. After, you must delete every element equal tonums[i] - 1ornums[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 ofnumsis at most20000.

Each elementnums[i]is an integer in the range[1, 10000].

2. Implementation

(1) DP

思路:

(1) 这题由于选择了一个数后，就必须把比这个数小1或比这个数大于1的数给删除，所以我们需要用一个方法可以表示所有连续数字，由于数字的范围在1 - 10000,所以直接建立一个size为10000的数组dp

(2) 对于每个数，我们只有两种操作, earn it or skip it, 如果我们要earn的话，就要跳过比当前数少1的数，如果skip当前的数，则取之前的数，所以可以得到状态转移方程:

dp[i] = Math.max(dp[i - 1], dp[i - 2] + dp[i]);

(3) 上面的dp有两层含义，在第一步初始化dp数组时，dp[i]表示数字i的值总和 比如数组[3, 3, 3]，dp[3] = 9。在第二步中我们要找到最大可以earn的值的时候，dp[i]此时的含义是在前[1-i]中，我们通过earn和delete的操作，可以得到的最大的值

class Solution {
    public int deleteAndEarn(int[] nums) {
        int n = 10001;
        int[] dp = new int[n];

        for (int num : nums) {
            dp[num] += num;
        }

        for (int i = 2; i < n; i++) {
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + dp[i]);
        }
        return dp[n - 1];
    }
}

(2) 空间优化

class Solution {
    public int deleteAndEarn(int[] nums) {
        int n = 10001;
        int[] dp = new int[n];

        for (int num : nums) {
            dp[num] += num;
        }

        int skip = 0, take = 0;
        for (int i = 0; i < n; i++) {
            int takei = skip + dp[i];
            int skipi = Math.max(skip, take);
            take = takei;
            skip = skipi;
        }
        return Math.max(skip, take);
    }
}

3. Time & Space Complexity

DP: 时间复杂度O(n), 空间复杂度O(1), 空间虽然是O(1)，但实际上size是10001，涵盖数的所有范围

空间优化: 时间复杂度O(n), 空间复杂度O(1)