LeetCode 1014 Best Sightseeing Pair DP

You are given an integer array values where values[i] represents the value of the ith sightseeing spot. Two sightseeing spots i and j have a distance j - i between them.

The score of a pair (i < j) of sightseeing spots is values[i] + values[j] + i - j: the sum of the values of the sightseeing spots, minus the distance between them.

Return the maximum score of a pair of sightseeing spots.

Solution

设 \(dp[j]\) 表示以 \(j\) 结尾的答案，转移方程：

\[dp[j] = \max_i(v[i]+v[j]+i-j)=v[j]-j+\max_i(v[i]+i) \]

因此只需要维护每个位置结尾的 \(\max_i(v[i]+i)\)

点击查看代码

class Solution {
private:
    int dp[50004];
    int MAX = -1;
    int add[50004];
public:
    int maxScoreSightseeingPair(vector<int>& values) {
        int n = values.size();
        if(n==2){return values[0]+values[1]-1;}
        for(int i=0;i<n;i++){
            if(i==0)add[i] = values[i]+i;
            else{
                add[i] = max(add[i-1],values[i]+i);
            }
        }
        
            // dp[j] = max_i(values[i]+values[j]+i-j)
            //       = values[j]-j+max_i(values[i]+i)
        for(int j=1;j<n;j++){
                dp[j] = values[j]-j+add[j-1];
                MAX = max(MAX, dp[j]);
        }
        return MAX;
        
    }
};