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  • LC 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: 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].
    Runtime: 4852 ms, faster than 0.72% of C++ online submissions for Delete and Earn.

    自以为做了一个区间DP,结果并不是这样做。

    //
    // Created by yuxi on 2019/1/22.
    //
    
    #include <vector>
    #include <unordered_map>
    #include <algorithm>
    using namespace std;
    
    class Solution {
    public:
      unordered_map<int,int> mp;
      unordered_map<int,int> memo;
      int deleteAndEarn(vector<int>& nums) {
        if(nums.empty()) return 0;
        for(int x : nums) mp[x]++;
        vector<int> keys;
        for(auto it = mp.begin(); it != mp.end(); it++) keys.push_back(it->first);
        sort(keys.begin(),keys.end());
        vector<vector<int>> dp(keys.size(), vector<int>(keys.size(),0));
        for(int k=0; k<keys.size(); k++) {
          for(int i=0; i<keys.size(); i++) {
            int j = i - k;
            if(j < 0) continue;
            if(i == j) {
              dp[j][i] = keys[i] * mp[keys[i]];
              continue;
            }
            for(int l = j; l < i; l++) {
              if(keys[l] == keys[l+1]-1) {
                if(l+1 == i) dp[j][i] = max(dp[j][i], max(dp[j][l],dp[i][i]));
                else dp[j][i] = max(dp[j][i], dp[j][l]+dp[l+2][i]);
              }else {
                dp[j][i] = max(dp[j][i], dp[j][l]+dp[l+1][i]);
              }
            }
          }
        }
        return dp[0][keys.size()-1];
      }
    };

    Runtime: 8 ms, faster than 87.68% of C++ online submissions for Delete and Earn.

    //
    // Created by yuxi on 2019/1/22.
    //
    
    #include <vector>
    #include <unordered_map>
    #include <algorithm>
    using namespace std;
    
    class Solution {
    public:
      unordered_map<int,int> mp;
      unordered_map<int,int> memo;
      int deleteAndEarn(vector<int>& nums) {
        if(nums.empty()) return 0;
        int maxval = 0;
        for(int x : nums) {
          maxval = max(maxval,x);
          mp[x]++;
        }
        vector<int> a(maxval+1, 0);
        for(auto it=mp.begin(); it != mp.end(); it++) {
          a[it->first] = it->second;
        }
        int prev = 0, cur = 0;
        for(int i=1; i<a.size(); i++) {
          cur = max(cur, i > 2 ? a[i] * i + a[i-2] : a[i] * i);
          a[i] = cur;
        }
        return cur;
      }
    };
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  • 原文地址:https://www.cnblogs.com/ethanhong/p/10301704.html
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