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python常用算法总结（下）

news 来源：原创 2025/5/14 6:46:10

python常用算法总结

最大子数组和（Kadane’s Algorithm）

最长公共子序列（Longest Common Subsequence）

编辑距离（Edit Distance）

背包问题（Knapsack Problem）

最长上升子序列（Longest Increasing Subsequence）

汉诺塔问题（Tower of Hanoi）

N皇后问题（N-Queens Problem）

Rat in a Maze

图的颜色问题（Graph Coloring Problem）

约瑟夫斯问题（Josephus Problem）

集合划分问题（Partition Problem）

字符串匹配（KMP算法）

python常用算法总结

最大子数组和（Kadane’s Algorithm）

Kadane’s算法用于查找具有最大和的连续子数组。

def kadane(arr):max_so_far = arr[0]max_ending_here = arr[0]for i in range(1, len(arr)):max_ending_here = max(arr[i], max_ending_here + arr[i])max_so_far = max(max_so_far, max_ending_here)return max_so_far# ⽰例
print(kadane([1, -3, 2, 1, -1]))

最长公共子序列（Longest Common Subsequence）

LCS用于查找两个字符串的最长公共子序列。

def lcs(X, Y):m = len(X)n = len(Y)L = [[None]*(n+1) for i in range(m+1)]for i in range(m+1):for j in range(n+1):if i == 0 or j == 0:L[i][j] = 0elif X[i-1] == Y[j-1]:L[i][j] = L[i-1][j-1] + 1else:L[i][j] = max(L[i-1][j], L[i][j-1])return L[m][n]# ⽰例
X = "AGGTAB"
Y = "GXTXAYB"
print(lcs(X, Y))

编辑距离（Edit Distance）

编辑距离用于计算两个字符串之间转换的最少操作数。

def edit_distance(str1, str2):m = len(str1)n = len(str2)dp = [[0 for x in range(n+1)] for x in range(m+1)]for i in range(m+1):for j in range(n+1):if i == 0:dp[i][j] = jelif j == 0:dp[i][j] = ielif str1[i-1] == str2[j-1]:dp[i][j] = dp[i-1][j-1]else:dp[i][j] = 1 + min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1])return dp[m][n]# ⽰例
str1 = "sunday"
str2 = "saturday"
print(edit_distance(str1, str2))

背包问题（Knapsack Problem）

背包问题用于在不超过背包容量的情况下选择物品以获得最大价值。

def knapSack(W, wt, val, n):K = [[0 for x in range(W+1)] for x in range(n+1)]for i in range(n+1):for w in range(W+1):if i == 0 or w == 0:K[i][w] = 0elif wt[i-1] <= w:K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]], K[i-1][w])else:K[i][w] = K[i-1][w]return K[n][W]# ⽰例
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print(knapSack(W, wt, val, n))

最长上升子序列（Longest Increasing Subsequence）

LIS用于查找最长的严格上升子序列。

def lis(arr):n = len(arr)lis = [1]*nfor i in range(1, n):for j in range(0, i):if arr[i] > arr[j] and lis[i] < lis[j] + 1:lis[i] = lis[j]+1maximum = 0for i in range(n):maximum = max(maximum, lis[i])return maximum# ⽰例
arr = [10, 22, 9, 33, 21, 50, 41, 60, 80]
print(lis(arr))

汉诺塔问题（Tower of Hanoi）

汉诺塔问题是经典的递归问题，涉及移动圆盘。

def tower_of_hanoi(n, from_rod, to_rod, aux_rod):if n == 1:print("Move disk 1 from rod", from_rod, "to rod", to_rod)returntower_of_hanoi(n-1, from_rod, aux_rod, to_rod)print("Move disk", n, "from rod", from_rod, "to rod", to_rod)tower_of_hanoi(n-1, aux_rod, to_rod, from_rod)# ⽰例
n = 4
tower_of_hanoi(n, 'A', 'C', 'B')

N皇后问题（N-Queens Problem）

N皇后问题涉及在N x N的棋盘上放置N个皇后，使得它们彼此之间不会攻击。

def print_solution(board):for i in board:print(" ".join(str(x) for x in i))def is_safe(board, row, col, n):for i in range(col):if board[row][i] == 1:return False9for i, j in zip(range(row, -1, -1), range(col, -1, -1)):if board[i][j] == 1:return Falsefor i, j in zip(range(row, n, 1), range(col, -1, -1)):if board[i][j] == 1:return Falsereturn Truedef solve_nq_util(board, col, n):if col >= n:return Truefor i in range(n):if is_safe(board, i, col, n):board[i][col] = 1if solve_nq_util(board, col + 1, n):return Trueboard[i][col] = 0return Falsedef solve_nq(n):board = [[0 for j in range(n)] for i in range(n)]if not solve_nq_util(board, 0, n):print("Solution does not exist")return Falseprint_solution(board)return True# ⽰例
n = 4
solve_nq(n)

Rat in a Maze

Rat in a Maze问题是经典的回溯问题，涉及找到迷宫的所有可能路径。

def is_safe(maze, x, y, n):if x >= 0 and x < n and y >= 0 and y < n and maze[x][y] == 1:return Truereturn Falsedef solve_maze_util(maze, x, y, sol, n):if x == n - 1 and y == n - 1 and maze[x][y] == 1:sol[x][y] = 1return Trueif is_safe(maze, x, y, n):if sol[x][y] == 1:return Falsesol[x][y] = 1if solve_maze_util(maze, x + 1, y, sol, n):return Trueif solve_maze_util(maze, x, y + 1, sol, n):return Trueif solve_maze_util(maze, x - 1, y, sol, n):return Trueif solve_maze_util(maze, x, y - 1, sol, n):return Truesol[x][y] = 0return Falsedef solve_maze(maze):n = len(maze)sol = [[0 for j in range(n)] for i in range(n)]if not solve_maze_util(maze, 0, 0, sol, n):print("Solution does not exist")return Falseprint_solution(sol)return True# ⽰例
maze = [[1, 0, 0, 0],[1, 1, 0, 1],[0, 1, 0, 0],[1, 1, 1, 1]
]
solve_maze(maze)50

图的颜色问题（Graph Coloring Problem）

图的颜色问题涉及为图的顶点着色，使得相邻顶点不具有相同颜色。

def is_safe(v, graph, color, c):for i in range(len(graph)):if graph[v][i] == 1 and color[i] == c:return Falsereturn Truedef graph_coloring_util(graph, m, color, v):if v == len(graph):return Truefor c in range(1, m + 1):if is_safe(v, graph, color, c):color[v] = cif graph_coloring_util(graph, m, color, v + 1):return Truecolor[v] = 0def graph_coloring(graph, m):color = [0] * len(graph)if not graph_coloring_util(graph, m, color, 0):print("Solution does not exist")return Falseprint("Solution exists: ", color)return True# ⽰例
graph = [[0, 1, 1, 1],[1, 0, 1, 0],[1, 1, 0, 1],[1, 0, 1, 0]
]
m = 3
graph_coloring(graph, m)

约瑟夫斯问题（Josephus Problem）

约瑟夫斯问题涉及找到最后一个幸存者的位置。

def josephus(n, k):if n == 1:return 0else:return (josephus(n - 1, k) + k) % n# ⽰例
n = 7
k = 3
print("The chosen place is ", josephus(n, k))

集合划分问题（Partition Problem）

集合划分问题涉及将给定的集合划分为两个子集，使得它们的和相等。

def is_subset_sum(arr, n, sum):subset = [[False for i in range(sum + 1)] for i in range(n + 1)]for i in range(n + 1):subset[i][0] = Truefor i in range(1, sum + 1):subset[0][i] = Falsefor i in range(1, n + 1):for j in range(1, sum + 1):if j < arr[i - 1]:subset[i][j] = subset[i - 1][j]if j >= arr[i - 1]:subset[i][j] = (subset[i - 1][j] or subset[i - 1][j - arr[i - 1]])return subset[n][sum]def find_partition(arr, n):sum = 0for i in range(n):sum += arr[i]if sum % 2 != 0:return Falsereturn is_subset_sum(arr, n, sum // 2)# ⽰例
arr = [3, 1, 5, 9, 12]
n = len(arr)
if find_partition(arr, n) == True:print("Can be divided into two subsets of equal sum")
else:print("Cannot be divided into two subsets of equal sum")

字符串匹配（KMP算法）

KMP算法用于在文本中查找给定的模式字符串。

def KMPSearch(pat, txt):M = len(pat)N = len(txt)lps = [0]*Mj = 0computeLPSArray(pat, M, lps)i = 0while i < N:if pat[j] == txt[i]:i += 1j += 1if j == M:print("Found pattern at index " + str(i-j))j = lps[j-1]elif i < N and pat[j] != txt[i]:if j != 0:j = lps[j-1]else:i += 1def computeLPSArray(pat, M, lps):27 length = 0lps[0] = 0i = 1while i < M:if pat[i] == pat[length]:length += 1lps[i] = lengthi += 1else:if length != 0:length = lps[length-1]else:lps[i] = 0i += 1# ⽰例
txt = "ABABDABACDABABCABAB"
pat = "ABABCABAB"
KMPSearch(pat, txt)