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下载的网站模板怎么改,怎么找到装修公司电话,wordpress中文模版,阿里云网站空间做商城流程法在SDET面试中的重要性‌ 软件测试工程师#xff08;SDET#xff09;不仅需验证功能#xff0c;还需编写高效、可靠的代码。LeetCode算法题是面试常见环节#xff0c;能评估候选人的问题解决能力和编码习惯。本文精选10道高频题#xff0c;均来自真实SDET面试题库#…法在SDET面试中的重要性‌软件测试工程师SDET不仅需验证功能还需编写高效、可靠的代码。LeetCode算法题是面试常见环节能评估候选人的问题解决能力和编码习惯。本文精选10道高频题均来自真实SDET面试题库涵盖基础数据结构与算法。每道题附Python和Java解法强调可读性和效率时间/空间复杂度。练习时建议从测试角度思考输入边界、异常处理和性能优化这将提升实际工作能力。‌1. Two Sum两数之和‌‌题目描述‌给定整数数组nums和目标值target返回数组中两数之和等于target的下标。假设每个输入有且仅有一个解且同一元素不可重复使用。‌解题思路‌使用哈希表存储元素值到索引的映射。遍历数组检查target - 当前值是否在哈希表中。若存在返回对应下标否则将当前值加入哈希表。时间复杂度O(n)空间O(n)。‌Python解法‌pythonCopy Code def two_sum(nums, target): num_map {} for i, num in enumerate(nums): complement target - num if complement in num_map: return [num_map[complement], i] num_map[num] i return [] # 无解处理题目保证有解但测试需考虑边界‌Java解法‌javaCopy Code import java.util.HashMap; import java.util.Map; public int[] twoSum(int[] nums, int target) { MapInteger, Integer numMap new HashMap(); for (int i 0; i nums.length; i) { int complement target - nums[i]; if (numMap.containsKey(complement)) { return new int[]{numMap.get(complement), i}; } numMap.put(nums[i], i); } throw new IllegalArgumentException(No solution); // 异常处理强化健壮性 }‌2. Valid Parentheses有效的括号‌‌题目描述‌给定字符串s包含括号(),[],{}判断是否有效开闭匹配且顺序正确。‌解题思路‌使用栈处理。遍历字符串遇开括号入栈遇闭括号则检查栈顶是否匹配。若栈空或失配则无效遍历结束栈空则有效。时间复杂度O(n)。‌Python解法‌pythonCopy Code def is_valid(s): stack [] mapping {): (, ]: [, }: {} for char in s: if char in mapping: # 闭括号 top stack.pop() if stack else # if mapping[char] ! top: return False else: # 开括号 stack.append(char) return not stack # 栈空则有效‌Java解法‌javaCopy Code import java.util.Stack; public boolean isValid(String s) { StackCharacter stack new Stack(); MapCharacter, Character mapping new HashMap(); mapping.put(), (); mapping.put(], [); mapping.put(}, {); for (char c : s.toCharArray()) { if (mapping.containsKey(c)) { // 闭括号 char top stack.isEmpty() ? # : stack.pop(); if (top ! mapping.get(c)) return false; } else { // 开括号 stack.push(c); } } return stack.isEmpty(); }‌3. Reverse Integer整数反转‌‌题目描述‌给定32位有符号整数x返回其数字反转形式。若反转后溢出返回0。‌解题思路‌循环取x的末位构建新整数。关键点处理溢出32位范围: -2^31 到 2^31-1。使用temp rev * 10 pop前检查是否超限。时间复杂度O(log n)。‌Python解法‌pythonCopy Code def reverse(x): rev 0 sign -1 if x 0 else 1 x abs(x) while x: pop x % 10 x // 10 # 检查溢出 if rev (2**31 - 1 - pop) // 10: return 0 rev rev * 10 pop return rev * sign‌Java解法‌javaCopy Code public int reverse(int x) { int rev 0; while (x ! 0) { int pop x % 10; x / 10; // 正数溢出检查 if (rev Integer.MAX_VALUE/10 || (rev Integer.MAX_VALUE/10 pop 7)) return 0; // 负数溢出检查 if (rev Integer.MIN_VALUE/10 || (rev Integer.MIN_VALUE/10 pop -8)) return 0; rev rev * 10 pop; } return rev; }‌4. Merge Two Sorted Lists合并两个有序链表‌‌题目描述‌合并两个升序链表返回新链表。新链表需保持升序。‌解题思路‌迭代法。创建哑节点简化边界处理。比较两链表当前节点将较小者链入新链表。时间复杂度O(nm)。‌Python解法‌pythonCopy Code class ListNode: def __init__(self, val0, nextNone): self.val val self.next next def merge_two_lists(l1, l2): dummy ListNode(-1) current dummy while l1 and l2: if l1.val l2.val: current.next l1 l1 l1.next else: current.next l2 l2 l2.next current current.next current.next l1 if l1 else l2 # 剩余部分直接链入 return dummy.next‌Java解法‌javaCopy Code public class ListNode { int val; ListNode next; ListNode(int x) { val x; } } public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ListNode dummy new ListNode(-1); ListNode current dummy; while (l1 ! null l2 ! null) { if (l1.val l2.val) { current.next l1; l1 l1.next; } else { current.next l2; l2 l2.next; } current current.next; } current.next (l1 ! null) ? l1 : l2; // 处理剩余节点 return dummy.next; }‌5. Valid Anagram有效的字母异位词‌‌题目描述‌给定两个字符串s和t判断t是否为s的字母异位词字母相同但顺序不同。‌解题思路‌统计字符频率。用数组或哈希表记录s的字符频次遍历t减频次。若某字符频次为负或s/t长度不等则无效。时间复杂度O(n)。‌Python解法‌pythonCopy Code def is_anagram(s, t): if len(s) ! len(t): return False freq [0] * 26 # 假设仅小写字母 for char in s: freq[ord(char) - ord(a)] 1 for char in t: index ord(char) - ord(a) freq[index] - 1 if freq[index] 0: return False return True‌Java解法‌javaCopy Code public boolean isAnagram(String s, String t) { if (s.length() ! t.length()) return false; int[] freq new int[26]; for (char c : s.toCharArray()) { freq[c - a]; } for (char c : t.toCharArray()) { freq[c - a]--; if (freq[c - a] 0) return false; } return true; }‌6. Best Time to Buy and Sell Stock买卖股票的最佳时机‌‌题目描述‌给定股价数组prices选择一天买后续一天卖求最大利润。若无法盈利返回0。‌解题思路‌贪心法。遍历数组记录历史最低价计算当前价与最低价的差值作为利润并更新最大利润。时间复杂度O(n)。‌Python解法‌pythonCopy Code def max_profit(prices): if not prices: return 0 min_price prices[0] max_profit 0 for price in prices: if price min_price: min_price price else: profit price - min_price if profit max_profit: max_profit profit return max_profit‌Java解法‌javaCopy Code public int maxProfit(int[] prices) { int minPrice Integer.MAX_VALUE; int maxProfit 0; for (int price : prices) { if (price minPrice) { minPrice price; } else if (price - minPrice maxProfit) { maxProfit price - minPrice; } } return maxProfit; }‌7. Binary Tree Inorder Traversal二叉树的中序遍历‌‌题目描述‌给定二叉树根节点返回中序遍历左-根-右结果。‌解题思路‌递归或迭代。迭代法用栈模拟先压入所有左节点再弹出访问转至右子树。时间复杂度O(n)。‌Python解法‌pythonCopy Code class TreeNode: def __init__(self, val0, leftNone, rightNone): self.val val self.left left self.right right def inorder_traversal(root): stack, result [], [] current root while current or stack: while current: # 压入所有左节点 stack.append(current) current current.left current stack.pop() result.append(current.val) current current.right # 转向右子树 return result‌Java解法‌javaCopy Code public class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val x; } } public ListInteger inorderTraversal(TreeNode root) { ListInteger res new ArrayList(); StackTreeNode stack new Stack(); TreeNode curr root; while (curr ! null || !stack.isEmpty()) { while (curr ! null) { stack.push(curr); curr curr.left; } curr stack.pop(); res.add(curr.val); curr curr.right; } return res; }‌8. String to Integer (atoi)字符串转换整数‌‌题目描述‌实现atoi函数将字符串转为整数。需处理前导空格、正负号、非数字字符和溢出。‌解题思路‌逐字符解析。跳过空格检查符号累积数字。溢出时返回边界值。时间复杂度O(n)。‌Python解法‌pythonCopy Code def my_atoi(s): s s.strip() if not s: return 0 sign 1 if s[0] in [, -]: if s[0] -: sign -1 s s[1:] num 0 for char in s: if not char.isdigit(): break digit int(char) # 检查溢出 if num (2**31 - 1 - digit) // 10: return 2zwnj;**31 - 1 if sign 1 else -2**zwnj;31 num num * 10 digit return sign * num‌Java解法‌javaCopy Code public int myAtoi(String s) { s s.trim(); if (s.isEmpty()) return 0; int sign 1, i 0; if (s.charAt(0) - || s.charAt(0) ) { sign (s.charAt(0) -) ? -1 : 1; i; } long num 0; // 用long防溢出 while (i s.length() Character.isDigit(s.charAt(i))) { int digit s.charAt(i) - 0; num num * 10 digit; if (num Integer.MAX_VALUE) { return (sign 1) ? Integer.MAX_VALUE : Integer.MIN_VALUE; } i; } return (int) num * sign; }‌9. Fibonacci Number斐波那契数‌‌题目描述‌计算第n个斐波那契数F(0)0, F(1)1, F(n)F(n-1)F(n-2)。‌解题思路‌动态规划。用两个变量迭代存储前两值避免递归重复计算。时间复杂度O(n)。‌Python解法‌pythonCopy Code def fib(n): if n 1: return n a, b 0, 1 for _ in range(2, n 1): a, b b, a b return b‌Java解法‌javaCopy Code public int fib(int n) { if (n 1) return n; int a 0, b 1; for (int i 2; i n; i) { int next a b; a b; b next; } return b; }‌10. Implement Queue using Stacks用栈实现队列‌‌题目描述‌仅用栈操作push, pop, peek, empty实现队列。‌解题思路‌双栈法。一个输入栈in_stack处理push输出栈out_stack处理pop/peek。当out_stack空时将in_stack元素倒入out_stack。时间复杂度均摊O(1)。‌Python解法‌pythonCopy Code class MyQueue: def __init__(self): self.in_stack [] self.out_stack [] def push(self, x): self.in_stack.append(x) def pop(self): if not self.out_stack: while self.in_stack: self.out_stack.append(self.in_stack.pop()) return self.out_stack.pop() def peek(self): if not self.out_stack: while self.in_stack: self.out_stack.append(self.in_stack.pop()) return self.out_stack[-1] def empty(self): return not (self.in_stack or self.out_stack)‌Java解法‌javaCopy Code import java.util.Stack; class MyQueue { private StackInteger inStack; private StackInteger outStack; public MyQueue() { inStack new Stack(); outStack new Stack(); } public void push(int x) { inStack.push(x); } public int pop() { if (outStack.isEmpty()) { while (!inStack.isEmpty()) { outStack.push(inStack.pop()); } } return outStack.pop(); } public int peek() { if (outStack.isEmpty()) { while (!inStack.isEmpty()) { outStack.push(inStack.pop()); } } return outStack.peek(); } public boolean empty() { return inStack.isEmpty() outStack.isEmpty(); } }‌总结与面试建议‌以上10道题覆盖SDET面试高频考点数组、字符串、链表、树、栈/队列和基础算法。练习时注意‌测试驱动思维‌为每个解法添加边界测试如空输入、极值。‌复杂度分析‌在代码注释中注明时间/空间复杂度展现优化意识。‌语言选择‌Python简洁适合快速实现Java强调类型安全面试中根据岗位要求调整。持续刷题并模拟面试精选文章边缘AI的测试验证挑战从云到端的质量保障体系重构编写高效Gherkin脚本的五大核心法则10亿条数据统计指标验证策略软件测试从业者的实战指南数据对比测试Data Diff工具的原理与应用场景