搜狐-算法-一面

1.自我介绍
2.yolo算法和fast cnn 的区别、yolo的网络结构
https://zhuanlan.zhihu.com/p/25236464
4.如何数据扩充
5.Unet结构
6.注意力机制怎么加的
7.特种筛选如何筛选的

8.编程题1
寻找两个正序数组的中位数
https://leetcode-cn.com/problems/median-of-two-sorted-arrays/

class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        n1 = len(nums1)
        n2 = len(nums2)
        if n1 > n2:
            return self.findMedianSortedArrays(nums2,nums1)
        k = (n1 + n2 + 1)//2
        left = 0
        right = n1
        while left < right :
            m1 = left +(right - left)//2
            m2 = k - m1
            if nums1[m1] < nums2[m2-1]:
                left = m1 + 1
            else:
                right = m1
        m1 = left
        m2 = k - m1 
        c1 = max(nums1[m1-1] if m1 > 0 else float("-inf"), nums2[m2-1] if m2 > 0 else float("-inf") )
        if (n1 + n2) % 2 == 1:
            return c1
        c2 = min(nums1[m1] if m1 < n1 else float("inf"), nums2[m2] if m2 <n2 else float("inf"))
        return (c1 + c2) / 2

9.编程题2
无重复字符的最长子串
https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        if not s:return 0
        left = 0
        lookup = set()
        n = len(s)
        max_len = 0
        cur_len = 0
        for i in range(n):
            cur_len += 1
            while s[i] in lookup:
                lookup.remove(s[left])
                left += 1
                cur_len -= 1
            if cur_len > max_len:max_len = cur_len
            lookup.add(s[i])
        return max_len