原题链接在这里：

题目：

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: 6Explanation: 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: 9Explanation: 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].

题解：

Sort the numbers into bucket. 

If you take the current bucket, you can't take next to it.

include[i] denotes max points earned by taking bucket i, include[i] = exclude[i-1] + i*buckets[i].

exclude[i] denotes max points earned by skipping bucket i, exclude[i] = Math.max(include[i-1], exclude[i-1]).

Time Complexity: O(nums.length + range). range = 10000.

Space: O(range).

AC Java: 

1 class Solution { 2     public int deleteAndEarn(int[] nums) { 3         if(nums == null || nums.length == 0){ 4             return 0; 5         } 6          7         int n = 10001; 8         int [] buckets = new int[n]; 9         for(int num : nums){10             buckets[num]++;11         }12         13         int in = 0;14         int ex = 0;15         for(int i = 0; i

类似.