以下是20个常见的算法竞赛题目及其解答示例： 

1.最大子序列和问题

题目：给定一个整数数组，找到一个具有最大和的连续子数组。 示例解答：

def max_subarray_sum(nums):max_sum = float('-inf')curr_sum = 0for num in nums:curr_sum = max(num, curr_sum + num)max_sum = max(max_sum, curr_sum)return max_sumnums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
print(max_subarray_sum(nums))  # 输出：6

2.最长递增子序列问题

题目：给定一个整数数组，找到其中最长的递增子序列的长度。 示例解答：

def longest_increasing_subsequence(nums):n = len(nums)dp = [1] * nfor i in range(1, n):for j in range(i):if nums[i] > nums[j]:dp[i] = max(dp[i], dp[j] + 1)return max(dp)nums = [10, 9, 2, 5, 3, 7, 101, 18]
print(longest_increasing_subsequence(nums))  # 输出：4

3.最小生成树问题

题目：给定一个带权重的无向图，找到一个最小生成树，使得所有边的权重之和最小。 示例解答：

from collections import defaultdict
from heapq import *def minimum_spanning_tree(graph):mst = []visited = set()start_node = list(graph.keys())[0]visited.add(start_node)edges = [(cost, start_node, to) for to, cost in graph[start_node]]heapify(edges)while edges:cost, frm, to = heappop(edges)if to not in visited:visited.add(to)mst.append((frm, to, cost))for next_to, next_cost in graph[to]:if next_to not in visited:heappush(edges, (next_cost, to, next_to))return mstgraph = defaultdict(list)
graph['A'] = [('B', 2), ('C', 3)]
graph['B'] = [('A', 2), ('C', 1), ('D', 1)]
graph['C'] = [('A', 3), ('B', 1), ('D', 2)]
graph['D'] = [('B', 1), ('C', 2)]
print(minimum_spanning_tree(graph))  # 输出：[('A', 'B', 2), ('B', 'C', 1), ('C', 'D', 2)]

4.最短路径问题

题目：给定一个带权重的有向图，找到从起点到终点的最短路径。 示例解答：

from collections import defaultdict
from heapq import *def shortest_path(graph, start, end):distances = {node: float('inf') for node in graph}distances[start] = 0heap = [(0, start)]while heap:curr_dist, curr_node = heappop(heap)if curr_dist > distances[curr_node]:continueif curr_node == end:breakfor neighbor, weight in graph[curr_node]:distance = curr_dist + weightif distance < distances[neighbor]:distances[neighbor] = distanceheappush(heap, (distance, neighbor))return distances[end]graph = defaultdict(list)
graph['A'] = [('B', 2), ('C', 3)]
graph['B'] = [('C', 1), ('D', 1)]
graph['C'] = [('D', 2)]
graph['D'] = []
print(shortest_path(graph, 'A', 'D'))  # 输出：3

5.背包问题

题目：给定一组物品的重量和价值，以及一个背包的容量，找到能够装入背包的物品的最大总价值。 示例解答：

def knapsack(weights, values, capacity):n = len(weights)dp = [[0] * (capacity + 1) for _ in range(n + 1)]for i in range(1, n + 1):for j in range(1, capacity + 1):if weights[i - 1] <= j:dp[i][j] = max(dp[i - 1][j], values[i - 1] + dp[i - 1][j - weights[i - 1]])else:dp[i][j] = dp[i - 1][j]return dp[n][capacity]weights = [1, 3, 4, 5]
values = [1, 4, 5, 7]
capacity = 7
print(knapsack(weights, values, capacity))  # 输出：9

6.图的拓扑排序问题

题目：给定一个有向无环图，对其进行拓扑排序。 示例解答：

from collections import defaultdictdef topological_sort(graph):visited = set()stack = []def dfs(node):visited.add(node)for neighbor in graph[node]:if neighbor not in visited:dfs(neighbor)stack.append(node)for node in graph:if node not in visited:dfs(node)return stack[::-1]graph = defaultdict(list)
graph['A'] = ['B', 'C']
graph['B'] = ['D']
graph['C'] = ['D']
graph['D'] = []
print(topological_sort(graph))  # 输出：['A', 'C', 'B', 'D']

7.两数之和问题

 给定一个整数数组和一个目标值，找出数组中和为目标值的两个数。示例解答：

def twoSum(nums, target):hashmap = {}for i, num in enumerate(nums):complement = target - numif complement in hashmap:return [hashmap[complement], i]hashmap[num] = ireturn []nums = [2, 7, 11, 15]
target = 9
print(twoSum(nums, target))  # 输出 [0, 1]

8.最小编辑距离问题

题目：给定两个字符串，找到将一个字符串转换为另一个字符串所需的最小编辑次数（插入、删除、替换）。 示例解答：

def min_edit_distance(word1, word2):m, n = len(word1), len(word2)dp = [[0] * (n + 1) for _ in range(m + 1)]for i in range(m + 1):dp[i][0] = ifor j in range(n + 1):dp[0][j] = jfor i in range(1, m + 1):for j in range(1, n + 1):if word1[i - 1] == word2[j - 1]:dp[i][j] = dp[i - 1][j - 1]else:dp[i][j] = min(dp[i - 1][j - 1], dp[i][j - 1], dp[i - 1][j]) + 1return dp[m][n]word1 = "horse"
word2 = "ros"
print(min_edit_distance(word1, word2))  # 输出：3

9.最长公共子序列问题

题目：给定两个字符串，找到它们的最长公共子序列的长度。 示例解答：

def longest_common_subsequence(text1, text2):m, n = len(text1), len(text2)dp = [[0] * (n + 1) for _ in range(m + 1)]for i in range(1, m + 1):for j in range(1, n + 1):if text1[i - 1] == text2[j - 1]:dp[i][j] = dp[i - 1][j - 1] + 1else:dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])return dp[m][n]text1 = "abcde"
text2 = "ace"
print(longest_common_subsequence(text1, text2))  # 输出：3

10.无重复字符的最长子串问题

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。示例解答：

def lengthOfLongestSubstring(s):max_len = 0start = 0char_map = {}for i, char in enumerate(s):if char in char_map and start <= char_map[char]:start = char_map[char] + 1else:max_len = max(max_len, i - start + 1)char_map[char] = ireturn max_lens = "abcabcbb"
print(lengthOfLongestSubstring(s))  # 输出 3

11.最长回文子串问题

题目：给定一个字符串，找到其中最长的回文子串。 示例解答：

def longest_palindrome(s):n = len(s)dp = [[False] * n for _ in range(n)]start, max_len = 0, 1for i in range(n):dp[i][i] = Truefor i in range(n - 1, -1, -1):for j in range(i + 1, n):if s[i] == s[j]:if j - i == 1 or dp[i + 1][j - 1]:dp[i][j] = Trueif j - i + 1 > max_len:start = imax_len = j - i + 1return s[start:start + max_len]s = "babad"
print(longest_palindrome(s))  # 输出："bab"

12.最小覆盖子串问题

给你一个字符串 s 、一个字符串 t ，请在字符串 s 里面找出包含 t 所有字符的最小子串。示例解答：

def minWindow(s, t):if len(s) < len(t):return ""target_map = {}for char in t:target_map[char] = target_map.get(char, 0) + 1required = len(target_map)formed = 0window_map = {}ans = float('inf'), None, Noneleft, right = 0, 0while right < len(s):char = s[right]window_map[char] = window_map.get(char, 0) + 1if char in target_map and window_map[char] == target_map[char]:formed += 1while left <= right and formed == required:char = s[left]if right - left + 1 < ans[0]:ans = (right - left + 1, left, right)window_map[char] -= 1if char in target_map and window_map[char] < target_map[char]:formed -= 1left += 1right += 1return "" if ans[0] == float('inf') else s[ans[1]: ans[2] + 1]s = "ADOBECODEBANC"
t = "ABC"
print(minWindow(s, t))  # 输出 "BANC"

13.单词拆分问题

给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict，判定 s 是否可以被空格拆分为一个或多个在字典中出现的单词。示例解答：

def wordBreak(s, wordDict):word_set = set(wordDict)dp = [False] * (len(s) + 1)dp[0] = Truefor i in range(1, len(s) + 1):for j in range(i):if dp[j] and s[j:i] in word_set:dp[i] = Truebreakreturn dp[-1]s = "leetcode"
wordDict = ["leet", "code"]
print(wordBreak(s, wordDict))  # 输出 True

14.最小基因变化问题

给定两个字符串 start 和 end，以及一个字符串列表 bank，每次可以将一个字母替换为另一个字母。只有在 bank 中的有效变化才算作一次基因变化。返回从 start 到 end 的最小基因变化次数，如果无法实现，则返回 -1。示例解答：

from collections import dequedef minMutation(start, end, bank):bank_set = set(bank)if end not in bank_set:return -1queue = deque([(start, 0)])while queue:gene, level = queue.popleft()if gene == end:return levelfor i in range(len(gene)):for char in 'ACGT':new_gene = gene[:i] + char + gene[i+1:]if new_gene in bank_set:bank_set.remove(new_gene)queue.append((new_gene, level + 1))return -1start = "AACCGGTT"
end = "AACCGGTA"
bank = ["AACCGGTA"]
print(minMutation(start, end, bank))  # 输出 1

15.任务调度器问题

给定一个用字符数组表示的 CPU 需要执行的任务列表。其中包含使用大写的 A - Z 字母表示的26 种不同种类的任务。任务可以以任意顺序执行，并且每个任务都可以在 1 个单位时间内执行完。在任何一个单位时间，CPU 可以完成一个任务，或者处于待命状态。然而，两个相同种类的任务之间必须有长度为整数 n 的冷却时间，因此至少有连续 n 个单位时间内 CPU 在执行不同的任务，或者在待命状态。你需要计算完成所有任务所需要的最短时间。示例解答：

from collections import Counterdef leastInterval(tasks, n):counter = Counter(tasks)max_freq = max(counter.values())max_count = sum(1 for count in counter.values() if count == max_freq)return max(len(tasks), (max_freq - 1) * (n + 1) + max_count)tasks = ["A", "A", "A", "B", "B", "B"]
n = 2
print(leastInterval(tasks, n))  # 输出 8

16.单词接龙问题

给定两个单词（beginWord 和 endWord）和一个字典 wordList，找到从 beginWord 到 endWord 的最短转换序列的长度。转换需遵循如下规则：每次转换只能改变一个字母；转换过程中的中间单词必须是字典中的单词。示例解答：

from collections import dequedef ladderLength(beginWord, endWord, wordList):word_set = set(wordList)if endWord not in word_set:return 0queue = deque([(beginWord, 1)])while queue:word, length = queue.popleft()if word == endWord:return lengthfor i in range(len(word)):for char in 'abcdefghijklmnopqrstuvwxyz':new_word = word[:i] + char + word[i+1:]if new_word in word_set:word_set.remove(new_word)queue.append((new_word, length + 1))return 0beginWord = "hit"
endWord = "cog"
wordList = ["hot", "dot", "dog", "lot", "log", "cog"]
print(ladderLength(beginWord, endWord, wordList))  # 输出 5

17.课程表问题

现在你总共有 n 门课需要选，记为 0 到 n-1。给定一个数组 prerequisites ，其中 prerequisites[i] = [ai, bi] ，表示在选修课程 ai 前必须选修课程 bi 。例如，先修课程 1 后才能选修课程 0 ，表示为 [0,1] 。请你判断是否可能完成所有课程的学习？示例解答：

def canFinish(numCourses, prerequisites):graph = [[] for _ in range(numCourses)]visited = [0] * numCoursesfor pair in prerequisites:x, y = pairgraph[x].append(y)def dfs(i):if visited[i] == -1:return Falseif visited[i] == 1:return Truevisited[i] = -1for j in graph[i]:if not dfs(j):return Falsevisited[i] = 1return Truefor i in range(numCourses):if not dfs(i):return Falsereturn TruenumCourses = 2
prerequisites = [[1, 0]]
print(canFinish(numCourses, prerequisites))  # 输出 True

18.二叉树的最近公共祖先问题

给定一个二叉树，找到该树中两个指定节点的最近公共祖先。示例解答：

class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef lowestCommonAncestor(root, p, q):if not root or root == p or root == q:return rootleft = lowestCommonAncestor(root.left, p, q)right = lowestCommonAncestor(root.right, p, q)if left and right:return rootreturn left if left else rightroot = TreeNode(3)
root.left = TreeNode(5)
root.right = TreeNode(1)
root.left.left = TreeNode(6)
root.left.right = TreeNode(2)
root.right.left = TreeNode(0)
root.right.right = TreeNode(8)
root.left.right.left = TreeNode(7)
root.left.right.right = TreeNode(4)
p = root.left
q = root.right
print(lowestCommonAncestor(root, p, q).val)  # 输出 3

19.二叉树的层序遍历问题

给定一个二叉树，返回其按层序遍历得到的节点值（从左到右，逐层访问）。示例解答：

class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef levelOrder(root):if not root:return []result = []queue = [root]while queue:level = []for _ in range(len(queue)):node = queue.pop(0)level.append(node.val)if node.left:queue.append(node.left)if node.right:queue.append(node.right)result.append(level)return resultroot = TreeNode(3)
root.left = TreeNode(9)
root.right = TreeNode(20)
root.right.left = TreeNode(15)
root.right.right = TreeNode(7)
print(levelOrder(root))  # 输出 [[3], [9, 20], [15, 7]]

20.二叉树的最大深度问题

给定一个二叉树，找出其最大深度。示例解答：

class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef maxDepth(root):if not root:return 0return max(maxDepth(root.left), maxDepth(root.right)) + 1root = TreeNode(3)
root.left = TreeNode(9)
root.right = TreeNode(20)
root.right.left = TreeNode(15)
root.right.right = TreeNode(7)
print(maxDepth(root))  # 输出 3