PDF编辑、屏幕录制与OCR识别:一站式效率工具的设计逻辑与实战应用
2026/8/23 2:36:14
这道2026年华为暑期实习AI方向的选择题,主要考察以下几个核心能力:
提示:华为算法题通常会有明确的输入输出规范,需要特别注意题目中的约束条件
这类题目通常要求对数组进行某种变换或计算。常见解法包括:
# 示例:移除有序数组中的重复元素(双指针解法) def removeDuplicates(nums): if not nums: return 0 slow = 0 for fast in range(1, len(nums)): if nums[fast] != nums[slow]: slow += 1 nums[slow] = nums[fast] return slow + 1常考题型包括:
// 示例:验证回文字符串 public boolean isPalindrome(String s) { int left = 0, right = s.length() - 1; while (left < right) { while (left < right && !Character.isLetterOrDigit(s.charAt(left))) left++; while (left < right && !Character.isLetterOrDigit(s.charAt(right))) right--; if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) { return false; } left++; right--; } return true; }public class Solution { public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { int complement = target - nums[i]; if (map.containsKey(complement)) { return new int[] { map.get(complement), i }; } map.put(nums[i], i); } throw new IllegalArgumentException("No two sum solution"); } }class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, int> hash; for (int i = 0; i < nums.size(); ++i) { auto it = hash.find(target - nums[i]); if (it != hash.end()) { return {it->second, i}; } hash[nums[i]] = i; } return {}; } };def twoSum(nums, target): hash_map = {} for i, num in enumerate(nums): if target - num in hash_map: return [hash_map[target - num], i] hash_map[num] = i return []注意:在线判题系统通常对IO有时间限制,大量数据输入时需要考虑IO效率
从O(n²)优化到O(nlogn):
从O(n)优化到O(1):
以"两数之和"问题为例:
选择依据:
基础巩固:
专项突破:
模拟实战:
初级阶段(1-2个月):
进阶阶段(1个月):
冲刺阶段(2周):