[游戏开发]#740 Delete and Earn

Description

You are given an integer array nums. You want to maximize the number of points you get by performing the following operation any number of times:

• Pick any nums[i] and delete it to earn nums[i] points.

Afterwards, you must delete every element equal to nums[i] - 1 and every element equal to nums[i] + 1.
Return the maximum number of points you can earn by applying the above operation some number of times.

Examples

Example 1:

Input: nums = [3,4,2]
Output: 6
Explanation: You can perform the following operations:

• Delete 4 to earn 4 points. Consequently, 3 is also deleted. nums = [2].
• Delete 2 to earn 2 points. nums = [].
  You earn a total of 6 points.

Example 2:
Input: nums = [2,2,3,3,3,4]
Output: 9
Explanation: You can perform the following operations:

• Delete a 3 to earn 3 points. All 2’s and 4’s are also deleted. nums = [3,3].
• Delete a 3 again to earn 3 points. nums = [3].
• Delete a 3 once more to earn 3 points. nums = [].
  You earn a total of 9 points.

Constraints:
1 <= nums.length <= 2 * 104
1 <= nums[i] <= 104

思路

这里有一个容易被忽视的点是，他并不是不能同时earn位置上相邻的两个数，而是不能同时earn数值上相邻的两个数
在知道了这点之后，其实就是一个简单的动态规划。

代码

class Solution {
    public int deleteAndEarn(int[] nums) {
        Arrays.sort(nums);
        int i = 0;
        int pre = Integer.MIN_VALUE;
        
        int[] answer = new int[nums.length];
        answer[0] = 0;
        int temp = nums[0];
        while(i < nums.length && nums[i] == temp){
            answer[0] += nums[i];
            i++;
        }
        if(i == nums.length)
            return answer[0];
        pre = nums[0];
        
        answer[1] = 0;
        temp = nums[i];
        while(i < nums.length && nums[i] == temp){
            answer[1] += nums[i];
            i++;
        }
        if(i == nums.length){
            if(nums[i - 1] == pre + 1)
                return Math.max(answer[0], answer[1]);
            else
                return answer[0] + answer[1];
        }
        if(nums[i - 1] == pre + 1)
            answer[1] = Math.max(answer[0], answer[1]);
        else
            answer[1] = answer[0] + answer[1];
        pre = nums[i - 1];   
        
        int step = 2;
        while(i < nums.length){
            int sum = 0;
            temp = nums[i];
            while(i < nums.length && nums[i] == temp){
                sum += nums[i];
                i++;
            }
            
            if(nums[i - 1] == pre + 1)
                answer[step] = Math.max(answer[step - 2] + sum, answer[step - 1]);
            else
                answer[step] = Math.max(answer[step - 2], answer[step - 1]) + sum;
            step++;
            pre = nums[i - 1];
        }
        
        return Math.max(answer[step - 1], answer[step - 2]);
    }
}