golang 解法, 执行用时: 104 ms, 内存消耗: 7.6 MB, 提交时间: 2023-06-06 09:24:12

func equalPairs(grid [][]int) (ans int) {
	for i := range grid {
		for j := range grid {
			ok := 1
			for k := range grid {
				if grid[i][k] != grid[k][j] {
					ok = 0
					break
				}
			}
			ans += ok
		}
	}
	return
}

java 解法, 执行用时: 37 ms, 内存消耗: 44.9 MB, 提交时间: 2023-06-06 09:23:51

class Solution {
    public int equalPairs(int[][] grid) {
        int n = grid.length;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                int ok = 1;
                for (int k = 0; k < n; ++k) {
                    if (grid[i][k] != grid[k][j]) {
                        ok = 0;
                        break;
                    }
                }
                ans += ok;
            }
        }
        return ans;
    }
}

python3 解法, 执行用时: 2292 ms, 内存消耗: 20.4 MB, 提交时间: 2023-06-06 09:23:39

class Solution:
    def equalPairs(self, grid: List[List[int]]) -> int:
        n = len(grid)
        ans = 0
        for i in range(n):
            for j in range(n):
                ans += all(grid[i][k] == grid[k][j] for k in range(n))
        return ans

python3 解法, 执行用时: 224 ms, 内存消耗: 20.5 MB, 提交时间: 2023-06-06 09:23:28

class Solution:
    def equalPairs(self, grid: List[List[int]]) -> int:
        g = [list(col) for col in zip(*grid)]
        return sum(row == col for row in grid for col in g)

python3 解法, 执行用时: 60 ms, 内存消耗: 18.5 MB, 提交时间: 2022-08-05 15:36:38

class Solution:
    def equalPairs(self, grid: List[List[int]]) -> int:
        # 用哈希表统计每行出现的次数，然后遍历列，累加哈希表中列出现的次数。
        cnt = Counter(tuple(row) for row in grid)
        return sum(cnt[col] for col in zip(*grid))

python3 解法, 执行用时: 7912 ms, 内存消耗: 18.6 MB, 提交时间: 2022-08-05 15:33:50

class Solution:
    def equalPairs(self, grid: List[List[int]]) -> int:
        # 遍历每一行每一列，对比
        from functools import reduce
        ans, n = 0, len(grid)
        for i in range(n):
            for j in range(n):
                if reduce(lambda x,y : x and y, [grid[i][k] == grid[k][j] for k in range(n)]):
                    ans += 1
        return ans