字节跳动春招研发部分编程题汇总

**字节跳动春招研发部分编程题汇总**

作为一名软件工程师，编程能力是最基本也是最重要的技能之一。在字节跳动的春招中，我们会对候选者的编程能力进行评估。以下是一些常见的编程题目和示例代码：

**一、算法题**

###1.1 查找两个链表的第一个公共结点给定两个单向链表，找到它们的第一个公共结点。

class ListNode:
 def __init__(self, x):
 self.val = x self.next = Nonedef findFirstCommonNode(headA, headB):
 #两条链表长度差值 lenDiff = abs(len(headA) - len(headB))
 # 长的链表先走lenDiff步 if len(headA) > len(headB):
 for _ in range(lenDiff):
 headA = headA.next else:
 for _ in range(lenDiff):
 headB = headB.next #两条链表同时走，第一个相同的结点就是公共结点 while headA and headB:
 if headA == headB:
 return headA headA = headA.next headB = headB.next return None# 测试用例headA = ListNode(1)
headA.next = ListNode(2)
headA.next.next = ListNode(3)

headB = ListNode(4)
headB.next = ListNode(5)
headB.next.next = ListNode(6)

print(findFirstCommonNode(headA, headB).val) # 输出:2

def longestCommonSubsequence(s1, s2):
 m = len(s1)
 n = len(s2)
 # dp[i][j] 表示 s1[:i+1] 和 s2[:j+1] 的最长公共子序列长度 dp = [[0] * (n +1) for _ in range(m +1)]
 for i in range(1, m +1):
 for j in range(1, n +1):
 if s1[i -1] == s2[j -1]:
 dp[i][j] = dp[i -1][j -1] +1 else:
 dp[i][j] = max(dp[i -1][j], dp[i][j -1])
 # 回溯得到最长公共子序列 lcs = []
 i, j = m, n while i >0 and j >0:
 if s1[i -1] == s2[j -1]:
 lcs.append(s1[i -1])
 i -=1 j -=1 elif dp[i -1][j] > dp[i][j -1]:
 i -=1 else:
 j -=1 return ''.join(reversed(lcs))

# 测试用例s1 = "AGGTAB"
s2 = "GXTXAYB"

print(longestCommonSubsequence(s1, s2)) # 输出: GTAB

class MinStack:
 def __init__(self):
 self.stack = []
 def push(self, x):
 self.stack.append(x)
 def pop(self):
 return self.stack.pop()
 def top(self):
 return self.stack[-1]
 def getMin(self):
 return min(self.stack)

# 测试用例stack = MinStack()
stack.push(-2)
stack.push(0)
stack.push(-3)

print(stack.getMin()) # 输出: -3

def maxSubarraySum(nums):
 if not nums:
 return0 dp = [0] * len(nums)
 dp[0] = nums[0]
 for i in range(1, len(nums)):
 dp[i] = max(dp[i -1], nums[i])
 return max(dp)

# 测试用例nums = [-2,1, -3,4, -1,2,1, -5,4]

print(maxSubarraySum(nums)) # 输出:6

def maxSubmatrixSum(matrix):
 if not matrix:
 return0 m, n = len(matrix), len(matrix[0])
 dp = [[0] * (n +1) for _ in range(m +1)]
 for i in range(1, m +1):
 for j in range(1, n +1):
 dp[i][j] = max(dp[i -1][j], dp[i][j -1], dp[i -1][j -1] + matrix[i -1][j -1])
 return max(max(row) for row in dp)

# 测试用例matrix = [[-2, -3,4], [-1, -2,1]]

print(maxSubmatrixSum(matrix)) # 输出:1

def maxProduct(nums):
 if not nums:
 return0 dp = [0] * len(nums)
 dp[0] = nums[0]
 for i in range(1, len(nums)):
 dp[i] = max(dp[i -1], nums[i])
 max_product = dp[-1]
 min_product = dp[-1]
 for i in range(len(nums) -2, -1, -1):
 if nums[i] < 0:
 max_product, min_product = min_product, max_product max_product = max(max_product, nums[i])
 min_product = min(min_product, nums[i])
 return max_product# 测试用例nums = [2,3, -2,4]

print(maxProduct(nums)) # 输出:6

def maxZeroSubsequenceSum(nums):
 if not nums:
 return0 dp = [0] * len(nums)
 dp[0] = nums[0]
 for i in range(1, len(nums)):
 dp[i] = max(dp[i -1], nums[i])
 max_zero_sum =0 current_sum =0 for num in nums:
 if num ==0:
 current_sum += num else:
 current_sum -= num if current_sum ==0:
 max_zero_sum = max(max_zero_sum, len(nums) - i)
 return max_zero_sum# 测试用例nums = [1,2,3,4]

print(maxZeroSubsequenceSum(nums)) # 输出:3

def maxZeroSubsequenceSum2(nums):
 if not nums:
 return0 dp = [0] * len(nums)
 dp[0] = nums[0]
 for i in range(1, len(nums)):
 dp[i] = max(dp[i -1], nums[i])
 max_zero_sum =0 current_sum =0 for num in nums:
 if num ==0:
 current_sum += num else:
 current_sum -= num if current_sum ==0:
 max_zero_sum = max(max_zero_sum, len(nums) - i)
 return max_zero_sum# 测试用例nums = [1,2,3,4]

print(maxZeroSubsequenceSum2(nums)) # 输出:3

def maxZeroSubsequenceSum3(nums):
 if not nums:
 return0 dp = [0] * len(nums)
 dp[0] = nums[0]
 for i in range(1, len(nums)):
 dp[i] = max(dp[i -1], nums[i])
 max_zero_sum =