题目：原题链接（中等）
标签：哈希表、二分查找
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N l o g M ) O(NlogM) O(NlogM) | O ( N ) O(N) O(N) | 80ms (78.41%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class Solution:
def smallestCommonElement(self, mat: List[List[int]]) -> int:
size1, size2 = len(mat), len(mat[0])
hashmap = {i: 0 for i in range(size1)}
now = [m[0] for m in mat]
while len(set(now)) > 1:
max_val = max(now)
for i in range(size1):
idx = bisect.bisect(mat[i], max_val, lo=hashmap[i])
# print(mat[i], max_val, "->", idx - 1)
if mat[i][idx - 1] == max_val:
hashmap[i] = idx - 1
elif idx >= size2:
return -1
else:
hashmap[i] = idx
now = [mat[i][hashmap[i]] for i in range(size1)]
return now[0]
