【LeetCode 740】 Delete and Earn

题目描述

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].

思路

和 198. House Robber 类似。
看完这句话才想出来。2333
把每个值对应的分数统计下来，然后遍历数值范围，当前数的最大分数取决于 -2个+当前值 和 -1个 的最大值。

代码

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        int n = nums.size();
        map<int, int> mp;
        int nmax = 0;
        for (int i=0; i<n; ++i) {
            mp[nums[i]] += nums[i];
            nmax = max(nmax, nums[i]);
        }
        
        vector<int> dp(nmax+1, 0);
        for (int i=1; i<=nmax; ++i) {
            dp[i] = max((i > 1 ? dp[i-2] : 0) + mp[i], dp[i-1]);
        }
        
        return *max_element(dp.begin(), dp.end());
    }
};