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].

My Solutions:

对每拿一个数字来说，可以拿到所有这个数字的总和，但是不能拿到和比它小一位或者第一位的数字。因此，这个问题类似于house robber，如果拿了i-1，就不能拿i

Time: We performed a radix sort instead, so our complexity is O(N+W), where W is the range of allowable values for nums[i].

Space : O(W), the size of our count.

class Solution {
    public int deleteAndEarn(int[] nums) {
        
        if (nums == null || nums.length == 0) return 0;
        
        /*优化的话可以找到最大值，用这个值作为dp长度
        int max = 0;
        for (int i : nums) max = Math.max(i, max);
        */
        
        int[] dp = new int[10001]; // or dp[max + 1]
        for (int i : nums) dp[i] += i;
        
        for (int i = 2; i < dp.length; i++) {
            dp[i] = Math.max(dp[i] + dp[i - 2], dp[i - 1]);
        }
        
        return dp[10000]; // or dp[max]
    }
}