leetcode笔记（六）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].

解题思路

看题目，可以意识到是动态规划类型的题目，但不知道怎么写迭代式子，就是相不清楚状态。所以，一开始心虚的按照自己的贪婪算法实现了下，结果就是代码极多，但结果不能对～

无奈，开始查阅资料，找到了一篇比较靠谱的博客。借鉴他的思路，自己努力写了下动态规划的实现。思路的关键在于：

取出一个数，其收益为数的频数 × 数的值。按照规则，取出一个，必然取出该值的所有数。
两个状态，取出当前数的最大收益（maxFetch），不取当前数的最大收益（maxNoFetch）。
初始状态：
- 　maxFetch = 0, maxNoFetch = 0;
当前状态与上一状态的关系
- 不取当前数，则为上一状态的最大值（max(prevMaxFetch, prevMaxNoFetch)）。
- 取出当前数，若数和上一状态的数关联（+/- 1），则为prevMaxNoFetch + 取出数的收益。否则，为max(prevMaxFetch, prevMaxNoFetch) + 取出数的收益。

示例代码

class Solution {

public:

    int deleteAndEarn(vector<int>& nums) {

        map<int, int> freqs;

        int size = nums.size();

        for(int i = ; i < size; i++)

        {

            int curr = nums[i];

            // remember frequences for curr

            if(freqs.find(curr) == freqs.end())

            {

                freqs[curr] = ;

            }

            else

            {

                freqs[curr] += ;

            }

        }

        int maxFetch = , maxNoFetch = ;

        int prevMaxFetch = , prevMaxNoFetch = ;

        map<int, int>::iterator prevChoice;

        map<int, int>::iterator currChoice;

        // calculate maximum according to previous status

        for(currChoice = freqs.begin(); currChoice != freqs.end(); ++currChoice)

        {

            // initiate

            if(currChoice == freqs.begin())

            {

                // get this number

                maxFetch = currChoice->first * currChoice->second;

                // do not get this number

                maxNoFetch = ;

            }

            // transferring

            else

            {

                prevMaxFetch = maxFetch;

                prevMaxNoFetch = maxNoFetch;

                // do not get the number

                maxNoFetch = max(prevMaxFetch, prevMaxNoFetch);

                // get this number

                if(currChoice->first == prevChoice -> first +  || currChoice->first == prevChoice -> first - ) {

                    // related -> must not fetch previous node

                    maxFetch = prevMaxNoFetch + currChoice->first * currChoice->second;

                }

                else

                {   // non related

                    maxFetch = maxNoFetch + currChoice->first * currChoice->second;

                }

            }

            prevChoice = currChoice;

        }

        return max(maxFetch, maxNoFetch);

    }

};

巴特西

leetcode笔记（六）740. Delete and Earn

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