测试评估
| 指标 | 结果 |
|---|---|
| Base 测试 | FAIL |
| Plus 测试 | FAIL |
| 代码长度 | 532 字符 |
模型生成的代码 （532 字符）
def maxAverageOfPath(matrix):
"""
Given a square matrix of size N*N, find a path from the top-left cell to the
bottom-right cell moving only right or down that yields the maximum average cost.
Since the number of cells visited on any such path is always (2*N - 1),
maximizing the average is equivalent to maximizing the total sum.
"""
if not matrix or not matrix[0]:
return 0.0
n = len(matrix)
# dp[i][j] will hold the maximum sum to reach cell (i, j)
dp = [[0] * n for _ in range(n)]Base 失败测试用例
[[[1, 2, 3], [6, 5, 4], [7, 3, 9]]]
Plus 失败测试用例
[[[0, 0, 0], [0, 0, 0], [0, 0, 0]]]