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feat: Revised the book (#978)
* Sync recent changes to the revised Word. * Revised the preface chapter * Revised the introduction chapter * Revised the computation complexity chapter * Revised the chapter data structure * Revised the chapter array and linked list * Revised the chapter stack and queue * Revised the chapter hashing * Revised the chapter tree * Revised the chapter heap * Revised the chapter graph * Revised the chapter searching * Reivised the sorting chapter * Revised the divide and conquer chapter * Revised the chapter backtacking * Revised the DP chapter * Revised the greedy chapter * Revised the appendix chapter * Revised the preface chapter doubly * Revised the figures
This commit is contained in:
@ -17,7 +17,7 @@ def random_access(nums: list[int]) -> int:
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# 请注意,Python 的 list 是动态数组,可以直接扩展
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# 为了方便学习,本函数将 list 看作是长度不可变的数组
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# 为了方便学习,本函数将 list 看作长度不可变的数组
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def extend(nums: list[int], enlarge: int) -> list[int]:
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"""扩展数组长度"""
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# 初始化一个扩展长度后的数组
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@ -34,12 +34,12 @@ def insert(nums: list[int], num: int, index: int):
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# 把索引 index 以及之后的所有元素向后移动一位
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for i in range(len(nums) - 1, index, -1):
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nums[i] = nums[i - 1]
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# 将 num 赋给 index 处元素
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# 将 num 赋给 index 处的元素
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nums[index] = num
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def remove(nums: list[int], index: int):
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"""删除索引 index 处元素"""
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"""删除索引 index 处的元素"""
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# 把索引 index 之后的所有元素向前移动一位
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for i in range(index, len(nums) - 1):
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nums[i] = nums[i + 1]
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@ -57,7 +57,7 @@ if __name__ == "__main__":
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n2 = ListNode(2)
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n3 = ListNode(5)
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n4 = ListNode(4)
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# 构建引用指向
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# 构建节点之间的引用
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n0.next = n1
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n1.next = n2
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n2.next = n3
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@ -22,7 +22,7 @@ if __name__ == "__main__":
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nums.clear()
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print("\n清空列表后 nums =", nums)
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# 尾部添加元素
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# 在尾部添加元素
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nums.append(1)
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nums.append(3)
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nums.append(2)
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@ -30,7 +30,7 @@ if __name__ == "__main__":
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nums.append(4)
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print("\n添加元素后 nums =", nums)
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# 中间插入元素
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# 在中间插入元素
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nums.insert(3, 6)
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print("\n在索引 3 处插入数字 6 ,得到 nums =", nums)
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@ -6,17 +6,17 @@ Author: Krahets (krahets@163.com)
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class MyList:
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"""列表类简易实现"""
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"""列表类"""
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def __init__(self):
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"""构造方法"""
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self._capacity: int = 10 # 列表容量
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self._arr: list[int] = [0] * self._capacity # 数组(存储列表元素)
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self._size: int = 0 # 列表长度(即当前元素数量)
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self._size: int = 0 # 列表长度(当前元素数量)
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self._extend_ratio: int = 2 # 每次列表扩容的倍数
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def size(self) -> int:
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"""获取列表长度(即当前元素数量)"""
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"""获取列表长度(当前元素数量)"""
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return self._size
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def capacity(self) -> int:
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@ -37,7 +37,7 @@ class MyList:
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self._arr[index] = num
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def add(self, num: int):
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"""尾部添加元素"""
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"""在尾部添加元素"""
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# 元素数量超出容量时,触发扩容机制
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if self.size() == self.capacity():
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self.extend_capacity()
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@ -45,7 +45,7 @@ class MyList:
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self._size += 1
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def insert(self, num: int, index: int):
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"""中间插入元素"""
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"""在中间插入元素"""
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if index < 0 or index >= self._size:
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raise IndexError("索引越界")
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# 元素数量超出容量时,触发扩容机制
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@ -87,7 +87,7 @@ class MyList:
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if __name__ == "__main__":
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# 初始化列表
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nums = MyList()
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# 尾部添加元素
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# 在尾部添加元素
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nums.add(1)
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nums.add(3)
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nums.add(2)
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@ -95,7 +95,7 @@ if __name__ == "__main__":
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nums.add(4)
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print(f"列表 nums = {nums.to_array()} ,容量 = {nums.capacity()} ,长度 = {nums.size()}")
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# 中间插入元素
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# 在中间插入元素
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nums.insert(6, index=3)
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print("在索引 3 处插入数字 6 ,得到 nums =", nums.to_array())
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@ -24,7 +24,7 @@ def backtrack(
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# 计算该格子对应的主对角线和副对角线
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diag1 = row - col + n - 1
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diag2 = row + col
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# 剪枝:不允许该格子所在列、主对角线、副对角线存在皇后
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# 剪枝:不允许该格子所在列、主对角线、副对角线上存在皇后
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if not cols[col] and not diags1[diag1] and not diags2[diag2]:
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# 尝试:将皇后放置在该格子
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state[row][col] = "Q"
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@ -41,8 +41,8 @@ def n_queens(n: int) -> list[list[list[str]]]:
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# 初始化 n*n 大小的棋盘,其中 'Q' 代表皇后,'#' 代表空位
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state = [["#" for _ in range(n)] for _ in range(n)]
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cols = [False] * n # 记录列是否有皇后
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diags1 = [False] * (2 * n - 1) # 记录主对角线是否有皇后
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diags2 = [False] * (2 * n - 1) # 记录副对角线是否有皇后
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diags1 = [False] * (2 * n - 1) # 记录主对角线上是否有皇后
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diags2 = [False] * (2 * n - 1) # 记录副对角线上是否有皇后
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res = []
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backtrack(0, n, state, res, cols, diags1, diags2)
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@ -29,7 +29,7 @@ def while_loop_ii(n: int) -> int:
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"""while 循环(两次更新)"""
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res = 0
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i = 1 # 初始化条件变量
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# 循环求和 1, 4, ...
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# 循环求和 1, 4, 10, ...
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while i <= n:
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res += i
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# 更新条件变量
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@ -14,7 +14,7 @@ def move(src: list[int], tar: list[int]):
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def dfs(i: int, src: list[int], buf: list[int], tar: list[int]):
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"""求解汉诺塔:问题 f(i)"""
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"""求解汉诺塔问题 f(i)"""
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# 若 src 只剩下一个圆盘,则直接将其移到 tar
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if i == 1:
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move(src, tar)
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@ -28,7 +28,7 @@ def dfs(i: int, src: list[int], buf: list[int], tar: list[int]):
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def solve_hanota(A: list[int], B: list[int], C: list[int]):
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"""求解汉诺塔"""
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"""求解汉诺塔问题"""
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n = len(A)
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# 将 A 顶部 n 个圆盘借助 B 移到 C
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dfs(n, A, B, C)
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@ -22,7 +22,7 @@ def backtrack(choices: list[int], state: int, n: int, res: list[int]) -> int:
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def climbing_stairs_backtrack(n: int) -> int:
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"""爬楼梯:回溯"""
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choices = [1, 2] # 可选择向上爬 1 或 2 阶
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choices = [1, 2] # 可选择向上爬 1 阶或 2 阶
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state = 0 # 从第 0 阶开始爬
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res = [0] # 使用 res[0] 记录方案数量
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backtrack(choices, state, n, res)
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@ -14,7 +14,7 @@ def coin_change_dp(coins: list[int], amt: int) -> int:
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# 状态转移:首行首列
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for a in range(1, amt + 1):
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dp[0][a] = MAX
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# 状态转移:其余行列
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# 状态转移:其余行和列
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for i in range(1, n + 1):
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for a in range(1, amt + 1):
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if coins[i - 1] > a:
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@ -62,7 +62,7 @@ def edit_distance_dp(s: str, t: str) -> int:
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dp[i][0] = i
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for j in range(1, m + 1):
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dp[0][j] = j
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# 状态转移:其余行列
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# 状态转移:其余行和列
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for i in range(1, n + 1):
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for j in range(1, m + 1):
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if s[i - 1] == t[j - 1]:
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@ -7,10 +7,10 @@ Author: Krahets (krahets@163.com)
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def knapsack_dfs(wgt: list[int], val: list[int], i: int, c: int) -> int:
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"""0-1 背包:暴力搜索"""
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# 若已选完所有物品或背包无容量,则返回价值 0
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# 若已选完所有物品或背包无剩余容量,则返回价值 0
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if i == 0 or c == 0:
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return 0
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# 若超过背包容量,则只能不放入背包
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# 若超过背包容量,则只能选择不放入背包
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if wgt[i - 1] > c:
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return knapsack_dfs(wgt, val, i - 1, c)
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# 计算不放入和放入物品 i 的最大价值
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@ -24,13 +24,13 @@ def knapsack_dfs_mem(
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wgt: list[int], val: list[int], mem: list[list[int]], i: int, c: int
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) -> int:
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"""0-1 背包:记忆化搜索"""
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# 若已选完所有物品或背包无容量,则返回价值 0
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# 若已选完所有物品或背包无剩余容量,则返回价值 0
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if i == 0 or c == 0:
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return 0
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# 若已有记录,则直接返回
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if mem[i][c] != -1:
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return mem[i][c]
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# 若超过背包容量,则只能不放入背包
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# 若超过背包容量,则只能选择不放入背包
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if wgt[i - 1] > c:
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return knapsack_dfs_mem(wgt, val, mem, i - 1, c)
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# 计算不放入和放入物品 i 的最大价值
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@ -55,7 +55,7 @@ def min_path_sum_dp(grid: list[list[int]]) -> int:
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# 状态转移:首列
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for i in range(1, n):
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dp[i][0] = dp[i - 1][0] + grid[i][0]
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# 状态转移:其余行列
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# 状态转移:其余行和列
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for i in range(1, n):
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for j in range(1, m):
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dp[i][j] = min(dp[i][j - 1], dp[i - 1][j]) + grid[i][j]
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@ -16,7 +16,7 @@ class GraphAdjList:
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def __init__(self, edges: list[list[Vertex]]):
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"""构造方法"""
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# 邻接表,key: 顶点,value:该顶点的所有邻接顶点
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# 邻接表,key:顶点,value:该顶点的所有邻接顶点
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self.adj_list = dict[Vertex, list[Vertex]]()
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# 添加所有顶点和边
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for edge in edges:
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@ -62,7 +62,7 @@ class GraphAdjMat:
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# 索引越界与相等处理
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if i < 0 or j < 0 or i >= self.size() or j >= self.size() or i == j:
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raise IndexError()
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# 在无向图中,邻接矩阵沿主对角线对称,即满足 (i, j) == (j, i)
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# 在无向图中,邻接矩阵关于主对角线对称,即满足 (i, j) == (j, i)
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self.adj_mat[i][j] = 1
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self.adj_mat[j][i] = 1
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@ -29,7 +29,7 @@ def graph_bfs(graph: GraphAdjList, start_vet: Vertex) -> list[Vertex]:
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# 遍历该顶点的所有邻接顶点
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for adj_vet in graph.adj_list[vet]:
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if adj_vet in visited:
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continue # 跳过已被访问过的顶点
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continue # 跳过已被访问的顶点
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que.append(adj_vet) # 只入队未访问的顶点
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visited.add(adj_vet) # 标记该顶点已被访问
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# 返回顶点遍历序列
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@ -19,7 +19,7 @@ def dfs(graph: GraphAdjList, visited: set[Vertex], res: list[Vertex], vet: Verte
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# 遍历该顶点的所有邻接顶点
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for adjVet in graph.adj_list[vet]:
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if adjVet in visited:
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continue # 跳过已被访问过的顶点
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continue # 跳过已被访问的顶点
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# 递归访问邻接顶点
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dfs(graph, visited, res, adjVet)
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@ -14,7 +14,7 @@ class Pair:
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class ArrayHashMap:
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"""基于数组简易实现的哈希表"""
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"""基于数组实现的哈希表"""
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def __init__(self):
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"""构造方法"""
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@ -75,7 +75,7 @@ class MaxHeap:
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# 判空处理
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if self.is_empty():
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raise IndexError("堆为空")
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# 交换根节点与最右叶节点(即交换首元素与尾元素)
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# 交换根节点与最右叶节点(交换首元素与尾元素)
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self.swap(0, self.size() - 1)
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# 删除节点
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val = self.max_heap.pop()
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@ -23,8 +23,8 @@ def binary_search(nums: list[int], target: int) -> int:
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def binary_search_lcro(nums: list[int], target: int) -> int:
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"""二分查找(左闭右开)"""
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# 初始化左闭右开 [0, n) ,即 i, j 分别指向数组首元素、尾元素+1
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"""二分查找(左闭右开区间)"""
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# 初始化左闭右开区间 [0, n) ,即 i, j 分别指向数组首元素、尾元素+1
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i, j = 0, len(nums)
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# 循环,当搜索区间为空时跳出(当 i = j 时为空)
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while i < j:
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@ -47,6 +47,6 @@ if __name__ == "__main__":
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index: int = binary_search(nums, target)
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print("目标元素 6 的索引 = ", index)
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# 二分查找(左闭右开)
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# 二分查找(左闭右开区间)
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index: int = binary_search_lcro(nums, target)
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print("目标元素 6 的索引 = ", index)
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@ -7,7 +7,7 @@ Author: Krahets (krahets@163.com)
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def two_sum_brute_force(nums: list[int], target: int) -> list[int]:
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"""方法一:暴力枚举"""
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# 两层循环,时间复杂度 O(n^2)
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# 两层循环,时间复杂度为 O(n^2)
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for i in range(len(nums) - 1):
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for j in range(i + 1, len(nums)):
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if nums[i] + nums[j] == target:
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@ -17,9 +17,9 @@ def two_sum_brute_force(nums: list[int], target: int) -> list[int]:
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def two_sum_hash_table(nums: list[int], target: int) -> list[int]:
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"""方法二:辅助哈希表"""
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# 辅助哈希表,空间复杂度 O(n)
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# 辅助哈希表,空间复杂度为 O(n)
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dic = {}
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# 单层循环,时间复杂度 O(n)
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# 单层循环,时间复杂度为 O(n)
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for i in range(len(nums)):
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if target - nums[i] in dic:
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return [dic[target - nums[i]], i]
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@ -12,7 +12,7 @@ def bucket_sort(nums: list[float]):
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buckets = [[] for _ in range(k)]
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# 1. 将数组元素分配到各个桶中
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for num in nums:
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# 输入数据范围 [0, 1),使用 num * k 映射到索引范围 [0, k-1]
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# 输入数据范围为 [0, 1),使用 num * k 映射到索引范围 [0, k-1]
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i = int(num * k)
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# 将 num 添加进桶 i
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buckets[i].append(num)
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@ -32,7 +32,7 @@ def heap_sort(nums: list[int]):
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sift_down(nums, len(nums), i)
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# 从堆中提取最大元素,循环 n-1 轮
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for i in range(len(nums) - 1, 0, -1):
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# 交换根节点与最右叶节点(即交换首元素与尾元素)
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# 交换根节点与最右叶节点(交换首元素与尾元素)
|
||||
nums[0], nums[i] = nums[i], nums[0]
|
||||
# 以根节点为起点,从顶至底进行堆化
|
||||
sift_down(nums, i, 0)
|
||||
|
||||
@ -10,7 +10,7 @@ class QuickSort:
|
||||
|
||||
def partition(self, nums: list[int], left: int, right: int) -> int:
|
||||
"""哨兵划分"""
|
||||
# 以 nums[left] 作为基准数
|
||||
# 以 nums[left] 为基准数
|
||||
i, j = left, right
|
||||
while i < j:
|
||||
while i < j and nums[j] >= nums[left]:
|
||||
@ -50,11 +50,11 @@ class QuickSortMedian:
|
||||
|
||||
def partition(self, nums: list[int], left: int, right: int) -> int:
|
||||
"""哨兵划分(三数取中值)"""
|
||||
# 以 nums[left] 作为基准数
|
||||
# 以 nums[left] 为基准数
|
||||
med = self.median_three(nums, left, (left + right) // 2, right)
|
||||
# 将中位数交换至数组最左端
|
||||
nums[left], nums[med] = nums[med], nums[left]
|
||||
# 以 nums[left] 作为基准数
|
||||
# 以 nums[left] 为基准数
|
||||
i, j = left, right
|
||||
while i < j:
|
||||
while i < j and nums[j] >= nums[left]:
|
||||
@ -84,7 +84,7 @@ class QuickSortTailCall:
|
||||
|
||||
def partition(self, nums: list[int], left: int, right: int) -> int:
|
||||
"""哨兵划分"""
|
||||
# 以 nums[left] 作为基准数
|
||||
# 以 nums[left] 为基准数
|
||||
i, j = left, right
|
||||
while i < j:
|
||||
while i < j and nums[j] >= nums[left]:
|
||||
@ -103,7 +103,7 @@ class QuickSortTailCall:
|
||||
while left < right:
|
||||
# 哨兵划分操作
|
||||
pivot = self.partition(nums, left, right)
|
||||
# 对两个子数组中较短的那个执行快排
|
||||
# 对两个子数组中较短的那个执行快速排序
|
||||
if pivot - left < right - pivot:
|
||||
self.quick_sort(nums, left, pivot - 1) # 递归排序左子数组
|
||||
left = pivot + 1 # 剩余未排序区间为 [pivot + 1, right]
|
||||
|
||||
@ -13,7 +13,7 @@ def digit(num: int, exp: int) -> int:
|
||||
|
||||
def counting_sort_digit(nums: list[int], exp: int):
|
||||
"""计数排序(根据 nums 第 k 位排序)"""
|
||||
# 十进制的位范围为 0~9 ,因此需要长度为 10 的桶
|
||||
# 十进制的位范围为 0~9 ,因此需要长度为 10 的桶数组
|
||||
counter = [0] * 10
|
||||
n = len(nums)
|
||||
# 统计 0~9 各数字的出现次数
|
||||
|
||||
@ -35,7 +35,7 @@ class LinkedListDeque:
|
||||
def push(self, num: int, is_front: bool):
|
||||
"""入队操作"""
|
||||
node = ListNode(num)
|
||||
# 若链表为空,则令 front, rear 都指向 node
|
||||
# 若链表为空,则令 front 和 rear 都指向 node
|
||||
if self.is_empty():
|
||||
self._front = self._rear = node
|
||||
# 队首入队操作
|
||||
|
||||
@ -20,7 +20,7 @@ if __name__ == "__main__":
|
||||
n3 = TreeNode(val=3)
|
||||
n4 = TreeNode(val=4)
|
||||
n5 = TreeNode(val=5)
|
||||
# 构建引用指向(即指针)
|
||||
# 构建节点之间的引用(指针)
|
||||
n1.left = n2
|
||||
n1.right = n3
|
||||
n2.left = n4
|
||||
|
||||
Reference in New Issue
Block a user