Leetcode: 1937. Maximum Number of Points with Cost

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Description

You are given an m x n integer matrix points (0-indexed). Starting with 0 points, you want to maximize the number of points you can get from the matrix.
To gain points, you must pick one cell in each row. Picking the cell at coordinates (r, c) will add points[r][c] to your score.
However, you will lose points if you pick a cell too far from the cell that you picked in the previous row. For every two adjacent rows r and r + 1 (where 0 <= r < m - 1), picking cells at coordinates (r, c1) and (r + 1, c2) will subtract abs(c1 - c2) from your score.
Return the maximum number of points you can achieve.
abs(x) is defined as:

x for x >= 0.
-x for x < 0.

Example

Input: points = [[1,2,3],[1,5,1],[3,1,1]]
Output: 9
Explanation:
The blue cells denote the optimal cells to pick, which have coordinates (0, 2), (1, 1), and (2, 0).
You add 3 + 5 + 3 = 11 to your score.
However, you must subtract abs(2 - 1) + abs(1 - 0) = 2 from your score.
Your final score is 11 - 2 = 9.

Tips

m == points.length
n == points[r].length
1 <= m, n <= 105
1 <= m * n <= 105
0 <= points[r][c] <= 105

分析

如果不需要 abs(c1 -c2)，那么这题将会比较简单。加上限制条件后，作一些转换可以将 abs 消除掉

#（r, c1) and (r + 1, c2) will subtract abs(c1 - c2)
1):   dp[r+1][c2] = max( dp[r][c1] + c2 -c 1) // 保证 c2 >= c1
2):  dp[r+1][c3] = max( dp[r][c1] + c1 -c3) //  保证 c1 >= c3

化简为
dp[r+1][c2] = c2 + max(dp[r][c1] - c1)     // c2 >= c1
dp[r+1][c3] = -c3 + max(dp[r][c1] + c1).  // c1 >=c3


class Solution(object):
    def maxPoints(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        N = len(points)
        M = len(points[0])       
        memo = points[0]
        
        for i in range(1, N):
            tmp = [float('-inf')] * M
            m = float('-inf')
            for j in range(M):
                m = max(m, memo[j]+j)
                tmp[j] = -j + m + points[i][j]
              
            m = float('-inf')
            for j in range(M-1, -1, -1):
                m = max(m, memo[j]-j)
                tmp[j] = max(tmp[j], points[i][j]+m +j)
            memo = tmp
        return max(memo)

总结

速度上战胜了 80 % 的提交， 提交 3 次通过

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