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| 1 | +/* |
| 2 | + * Copyright (C) 2023, Saul Lawliet <october dot sunbathe at gmail dot com> |
| 3 | + * All rights reserved. |
| 4 | + * |
| 5 | + * 直观感觉这道题不应该算中等难度, 解题的过程中有两个问题 |
| 6 | + * 1. hash越界 |
| 7 | + * 解决方法: 对10000007取模 |
| 8 | + * 2. hash一致但数组不相等 |
| 9 | + * 解决方法: hash一致的情况也继续对比数组是否相等 |
| 10 | + */ |
| 11 | + |
| 12 | +#include "c/data-structures/array.h" |
| 13 | +#include "c/test.h" |
| 14 | + |
| 15 | +int equalPairs(int **grid, int gridSize, int *gridColSize) { |
| 16 | + int *rowHash = malloc(sizeof(int) * gridSize); |
| 17 | + int *colHash = malloc(sizeof(int) * gridSize); |
| 18 | + |
| 19 | + for (int i = 0; i < gridSize; i++) { |
| 20 | + rowHash[i] = 1; |
| 21 | + colHash[i] = 1; |
| 22 | + } |
| 23 | + |
| 24 | + for (int i = 0; i < gridSize; i++) { |
| 25 | + for (int j = 0; j < gridSize; j++) { |
| 26 | + rowHash[i] = (rowHash[i] * 31 + grid[i][j]) % 10000007; |
| 27 | + colHash[i] = (colHash[i] * 31 + grid[j][i]) % 10000007; |
| 28 | + } |
| 29 | + } |
| 30 | + |
| 31 | + int rt = 0; |
| 32 | + for (int i = 0; i < gridSize; i++) { |
| 33 | + for (int j = 0; j < gridSize; j++) { |
| 34 | + if (rowHash[i] == colHash[j]) { |
| 35 | + // hash 一致, 也要对比每一项 |
| 36 | + for (int k = 0; k < gridSize; k++) { |
| 37 | + if (grid[i][k] != grid[k][j]) break; |
| 38 | + if (k == gridSize - 1) rt++; |
| 39 | + } |
| 40 | + } |
| 41 | + } |
| 42 | + } |
| 43 | + |
| 44 | + free(colHash); |
| 45 | + free(rowHash); |
| 46 | + |
| 47 | + return rt; |
| 48 | +} |
| 49 | + |
| 50 | +void test(int except, char *grid) { |
| 51 | + arrayEntry *e = arrayParse2D(grid, ARRAY_INT); |
| 52 | + EXPECT_EQ_INT(except, equalPairs(arrayValue(e), arrayRow(e), arrayCols(e))); |
| 53 | + arrayFree(e); |
| 54 | +} |
| 55 | + |
| 56 | +int main(void) { |
| 57 | + test(1, "[[3,2,1],[1,7,6],[2,7,7]]"); |
| 58 | + test(3, "[[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]"); |
| 59 | + test(0, "[[2,1],[3,32]]"); |
| 60 | + |
| 61 | + return testOutput(); |
| 62 | +} |
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