网站如何让百度抓取,做网站要备案,怎么制作公众号封面图,注册公司地址可以用家庭地址1.两数之和 (Two Sum) 给定一个整数数组 nums 和一个目标值 target#xff0c;请你在该数组中找出和为目标值的那两个整数#xff0c;并返回它们的数组下标。 Given an array of integers nums and a target value target, find the two integers in the array that sum up t…1.两数之和 (Two Sum) 给定一个整数数组 nums 和一个目标值 target请你在该数组中找出和为目标值的那两个整数并返回它们的数组下标。 Given an array of integers nums and a target value target, find the two integers in the array that sum up to the target value and return their array indices.
解释可以使用哈希表来存储已经遍历过的数字及其索引这样可以在 O(1) 时间内查找 target - nums[i] 是否存在。 Explanation: You can use a hash table to store the numbers and their indices that have been traversed, so you can find if target - nums[i] exists in O(1) time.
伪代码
hash_table {}
for i 0 to len(nums) - 1complement target - nums[i]if complement in hash_tablereturn [hash_table[complement], i]hash_table[nums[i]] i
return []2.有效的括号 (Valid Parentheses) 给定一个只包括 ‘(’‘)’‘{’‘}’‘[’‘]’ 的字符串判断字符串是否有效。 Given a string containing just the characters ‘(’, ‘)’, ‘{’, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.
解释使用栈来跟踪括号的匹配左括号入栈右括号时检查栈顶元素是否匹配。 Explanation: Use a stack to track the matching of parentheses. Push left parentheses onto the stack, and when encountering a right parenthesis, check if the top element of the stack matches.
伪代码
stack []
for each char in stringif char is left parenthesisstack.push(char)else if stack is not empty and top of stack matches charstack.pop()elsereturn false
return stack is empty3.合并两个有序链表 (Merge Two Sorted Lists) 将两个升序链表合并为一个新的升序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 Merge two sorted linked lists into a new sorted linked list and return it. The new list is formed by splicing together the nodes of the given two lists.
解释可以使用一个新的链表来逐个比较两个链表的节点选择较小的节点添加到新链表中。 Explanation: You can use a new linked list to compare the nodes of the two lists one by one, and add the smaller node to the new list.
伪代码
dummy new Node(0)
current dummy
while list1 is not null and list2 is not nullif list1.val list2.valcurrent.next list1list1 list1.nextelsecurrent.next list2list2 list2.nextcurrent current.next
if list1 is not nullcurrent.next list1
elsecurrent.next list2
return dummy.next4.盛最多水的容器 (Container With Most Water) 给定 n 个非负整数 a1a2…an每个数代表坐标中的一个点 (i, ai)。在坐标内画 n 条垂直线垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0)。找出其中的两条线使得它们与 x 轴共同构成的容器可以容纳最多的水。 Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). Draw n vertical lines such that the two endpoints of line i are at (i, ai) and (i, 0). Find two lines, which together with the x-axis form a container, such that the container contains the most water.
解释使用两个指针分别指向数组的开始和结束计算当前的容量并逐步向中间移动较短的线以寻找可能的更大容量。 Explanation: Use two pointers pointing to the start and end of the array, calculate the current capacity, and gradually move the shorter line towards the middle to find a potentially larger capacity.
伪代码
left 0
right len(height) - 1
max_water 0
while left rightwidth right - leftmax_water max(max_water, min(height[left], height[right]) * width)if height[left] height[right]leftelseright--
return max_water5.无重复字符的最长子串 (Longest Substring Without Repeating Characters) 给定一个字符串请你找出其中不含有重复字符的最长子串的长度。 Given a string, find the length of the longest substring without repeating characters.
解释可以使用滑动窗口的方法用一个哈希集合来记录窗口内的字符当遇到重复字符时移动窗口的开始位置。 Explanation: You can use the sliding window method and a hash set to record the characters within the window. When a repeated character is encountered, move the start position of the window.
伪代码
start 0
max_length 0
char_set set()
for i 0 to len(string) - 1while string[i] in char_setchar_set.remove(string[start])startchar_set.add(string[i])max_length max(max_length, i - start 1)
return max_length6.三数之和 (3Sum) 给你一个包含 n 个整数的数组 nums判断 nums 中是否存在三个元素 abc 使得 a b c 0请你找出所有满足条件且不重复的三元组。 Given an array nums of n integers, are there elements a, b, c in nums such that a b c 0? Find all unique triplets in the array which gives the sum of zero.
解释可以先对数组排序然后使用一个固定的指针遍历数组对于每个元素使用两个指针在剩余部分进行搜索找到满足条件的三元组。 Explanation: You can first sort the array and then use a fixed pointer to traverse the array. For each element, use two pointers to search in the remaining part to find triplets that meet the condition.
伪代码
sort(nums)
result []
for i 0 to len(nums) - 3if i 0 and nums[i] nums[i - 1]continueleft i 1right len(nums) - 1while left rightsum nums[i] nums[left] nums[right]if sum 0result.add([nums[i], nums[left], nums[right]])while left right and nums[left] nums[left 1]leftwhile left right and nums[right] nums[right - 1]right--leftright--else if sum 0leftelseright--
return result7.最接近的三数之和 (3Sum Closest) 给定一个包括 n 个整数的数组 nums 和一个目标值 target。找出 nums 中的三个整数使得它们的和与 target 最接近。返回这三个数的和。 Given an array nums of n integers and a target value target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers.
解释和三数之和类似先对数组排序然后使用一个固定的指针遍历数组对于每个元素使用两个指针在剩余部分进行搜索记录最接近的和。 Explanation: Similar to 3Sum, first sort the array, then use a fixed pointer to traverse the array. For each element, use two pointers to search in the remaining part and record the closest sum.
伪代码
sort(nums)
closest_sum inf
for i 0 to len(nums) - 3if i 0 and nums[i] nums[i - 1]continueleft i 1right len(nums) - 1while left rightsum nums[i] nums[left] nums[right]if abs(sum - target) abs(closest_sum - target)closest_sum sumif sum targetleftelse if sum targetright--elsereturn sum
return closest_sum
8.删除链表的倒数第N个节点 (Remove Nth Node From End of List) 给定一个链表删除链表的倒数第 n 个节点并且返回链表的头结点。 Given a linked list, remove the nth node from the end of the list and return its head.
解释可以使用两个指针第一个指针先移动 n 步然后两个指针同时移动直到第一个指针到达末尾这时第二个指针指向的就是需要删除的节点的前一个节点。 Explanation: You can use two pointers, move the first pointer n steps first, then move both pointers at the same time until the first pointer reaches the end, at which point the second pointer points to the node before the one to be deleted.
伪代码
dummy new Node(0)
dummy.next head
first dummy
second dummy
for i 0 to nfirst first.next
while first.next is not nullfirst first.nextsecond second.next
second.next second.next.next
return dummy.next9.回文数 (Palindrome Number) 判断一个整数是否是回文数。回文数是指正序从左向右和倒序从右向左读都是一样的整数。 Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
解释可以将整数转换为字符串然后使用双指针法从两端向中间比较字符是否相同。 Explanation: You can convert the integer to a string and then use the two-pointer technique to compare characters from both ends towards the middle to see if they are the same.
伪代码
str_num str(num)
left 0
right len(str_num) - 1
while left rightif str_num[left] ! str_num[right]return falseleftright--
return true10.整数反转 (Reverse Integer) 给出一个 32 位的有符号整数你需要将这个整数中每位上的数字进行反转。 Given a 32-bit signed integer, reverse digits of an integer.
解释可以通过循环取出整数的最后一位数字然后添加到新整数的末尾注意处理整数溢出的情况。 Explanation: You can reverse the digits of an integer by repeatedly popping the last digit of the integer and appending it to the end of a new integer, being careful to handle integer overflow.
伪代码
rev 0
while num ! 0pop num % 10num / 10if rev INT_MAX/10 or (rev INT_MAX / 10 and pop 7)return 0if rev INT_MIN/10 or (rev INT_MIN / 10 and pop -8)return 0rev rev * 10 pop
return rev
