家居网站关键词怎么做,网页制作培训多钱,网站建设会出现哪些问题,建设通查询设通网站以下是20个常见的算法竞赛题目及其解答示例#xff1a;
1.最大子序列和问题
题目#xff1a;给定一个整数数组#xff0c;找到一个具有最大和的连续子数组。 示例解答#xff1a;
def max_subarray_sum(nums):max_sum float(-inf)curr_sum 0for num in nums:curr_sum…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[i1:]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[i1:]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, val0, leftNone, rightNone):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, val0, leftNone, rightNone):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, val0, leftNone, rightNone):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
