LeetCode740

EX740

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.

Example1:

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.

Example2:

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].

Solution:

dp[i] 表示轮到i时，现有的point
以例二为例：

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对于值为i的元素，要么抛弃它，要么选择它
选择它以为着抛弃i-1和i+1, dp[i] = 所有i的和 + dp[i+2]
而抛弃它意味着dp[i] = dp[i+1]

class Solution {public:    int deleteAndEarn(vector<int>& nums) {        int len = nums.size();        vector<int> num(10001, 0);        vector<int> dp(10003, 0);        for (int i = 0; i < len; i++) {            num[nums[i]] += nums[i];        }        for (int i = 10000; i >= 0; i--) {            dp[i] = max(num[i]+dp[i+2], dp[i+1]);        }        return dp[0];    }};