leetcode 740. Delete and Earn题解

Posted 2021-01-05 J1ac

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】这道题初看觉得就是求每个连续区间奇偶分别求和的大者，然后仔细一看才发现是和之前HouseRobber一样的题目。

我自己写的时候把每个连续区间分离出来然后分别调用HouseRobber的DP函数最后求和，速度慢了一点。

看了网上其他解答才发现整个数字区间可以直接表示出来，然后调用一次HouseRobber的DP算法就好了。

自己的代码：

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        map<int,int> d;
        for(auto i:nums){
            d[i] += 1;
        }
        int res = 0;
        auto it = d.begin();
        while(it != d.end()){
            vector<int> temp;
            while(1){
                temp.push_back(it->first * it->second);
                auto pre = it->first;
                if(++it == d.end() || pre != it->first - 1) break;
            }
            res += maxGet(temp);
        }
        return res;
    }
    
    int maxGet(vector<int> & num){
        int s = num.size();
        vector<int> dp(s+1,0);
        dp[1] = num[0];
        for(int i = 2; i<=s; ++i){
            dp[i] = max(dp[i-1],dp[i-2] + num[i-1]);
        }
        return dp[s];
    }
};

更快的代码：

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