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:
nums is at most 20000.nums[i] is an integer in the range [1, 10000].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：
class Solution { public: int deleteAndEarn(vector<int>& nums) { map<int, int> freqs; int size = nums.size(); for(int i = 0; i < size; i++) { int curr = nums[i]; // remember frequences for curr if(freqs.find(curr) == freqs.end()) { freqs[curr] = 1; } else { freqs[curr] += 1; } } int maxFetch = 0, maxNoFetch = 0; int prevMaxFetch = 0, prevMaxNoFetch = 0; 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 = 0; } // transferring else { prevMaxFetch = maxFetch; prevMaxNoFetch = maxNoFetch; // do not get the number maxNoFetch = max(prevMaxFetch, prevMaxNoFetch); // get this number if(currChoice->first == prevChoice -> first + 1 || currChoice->first == prevChoice -> first - 1) { // 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); } };
leetcode笔记（六）740. Delete and Earn
原文：https://www.cnblogs.com/niuxu18/p/note_leetcode_6.html