leetcode weekly contest 61 （ 740. Delete and Earn ）

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: 6Explanation: 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: 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].

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题目

说我们可以从一堆数中挑选一个数 ， 然后删除比这个数大1和小1的数，然后得到挑选这个数的分
问最后最多可以拿多少分

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分析

首先如果我们有很多数这里数有重复那么我们一旦拿了其中一个数，就一定会拿完
因为当我们拿了一个数后，它已经没有这个数+1 和这个数-1 了，如果我们不拿完这个数，那么一定会剩下来

那么这个问题就简化了

接着是一个决策问题

很容易看到每一次都会作出选择是否拿这个数

并且最大化利益 opt = max
按顺序排序数
定义dp[i]代表拿这个数所能获得的最大利益

状态转移 如果前一项等于我们这个值-1

那么我们拿前一项的前一项的dp[i-2]+这项获得的利益 和前一项dp[i-1]比较

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代码

class Solution {public:    int deleteAndEarn(vector<int>& nums) {        if( nums.size() == 0 )            return 0 ;        map< int , int > hash ;        for( int i=0 ; i<nums.size() ; i++){            hash[nums[i]]++;        }        vector< int > vals;        for( auto it = hash.begin() ; it!= hash.end() ; it++){            vals.push_back( it->first  );        }        int ans = 0 ;        vector< int  > dp ;        dp.resize( vals.size() , 0 ) ;        dp[0] = vals[0]*hash[vals[0]];        for( int i=1 ; i<vals.size() ; i++){            if( vals[i-1] + 1 == vals[i] ){                if( i-2 >=0 ){                    dp[i] = max( dp[i-2] + vals[i]*hash[vals[i]] , dp[i-1] ) ;                }                else{                    dp[i] = max( vals[i]*hash[vals[i]] , dp[i-1] ) ;                }            }            else{                dp[i] = dp[i-1]+vals[i]*hash[vals[i]];            }        }        return dp.back() ;    }};

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时间复杂度O(n∗logn) 因为我们排序了可以通过hash降到O（n）

空间复杂度 O(n)