[LeetCode]740. Delete and Earn

descrption

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].

解题思路

这道题大致意思是让随意挑选数字nums[i]删除它,得到nums[i]分，但是同时必须删除数组中所有等于nums[i]-1和nums[i]+1的元素，直到数组中的所有元素都被删除。这里用到动态规划求解，可以用数组times[]记录数组num[]中每个数字出现的次数，即times[i]就是数组num[]中i出现的次数。按照题目要求，如果我们选择数字i，那么相应的就不能选择i-1和i+1，这里可以定义状态转移方程：

用points[i]记录遍历到数字i时的最大得分 ：
points[i] = max(points[i-2]+times[i]*i, points[i-1])

代码如下

class Solution {
public:
int deleteAndEarn(vector<int>& nums) {
vector<int> times(10001, 0);
//count appear times for every num in nums
for (int i = 0; i < nums.size(); i++) {
times[nums[i]]++;
}
vector<int> points(10001, 0);
points[1] = times[1]*1;
for (int i = 2; i <= points.size(); i++) {
points[i] = max(points[i-1], points[i-2]+times[i]*i);
}
return points[10000];
}
};

原题地址

如有错误请指出，谢谢！