Python算法数据结构

挑选自《图解算法与数据结构》中较为简单的部分题目。
原文地址如下：
https://leetcode.cn/leetbook/detail/illustration-of-algorithm/

1 搜索与回溯算法

1.1 剑指 Offer 12. 矩阵中的路径

dfs + 回溯

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        def dfs(i, j, k):
            if not 0 <= i < len(board) or not 0 <= j < len(board[0]) or board[i][j] != word[k]:
                return False
            if k == len(word) - 1:
                return True
            board[i][j] = ''
            res = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)
            board[i][j] = word[k]
            return res
        for i in range(len(board)):
            for j in range(len(board[0])):
                if dfs(i, j, 0):
                    return True
        return False

1.2 机器人的运动范围

dfs + 回溯

class Solution:
    def movingCount(self, m: int, n: int, k: int) -> int:
        vis = set()
        def sum1(k):
            b = 0
            for a in str(k):
                b += int(a)
            return b
        def dfs(i, j):
            if i >= m or j >= n or k < sum1(i) + sum1(j) or (i, j) in vis:
                return 0
            vis.add((i, j))
            return 1 + dfs(i + 1, j) + dfs(i, j + 1)
        return dfs(0, 0)

1.3 树的子结构

真头疼啊，一点一点想想吧。
首先是寻找匹配节点，即T.val == B.val的时候。
A遍历为空，返回False。
B为空，返回为False。
匹配到B，返回True。
然后匹配B，左右匹配成功，B为空，返回True。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSubStructure(self, A: TreeNode, B: TreeNode) -> bool:
        def dfs(A, B):
            if not A or not B:
                return False
            if B and A.val == B.val and check(A, B):
                return True
            return dfs(A.left, B) or dfs(A.right, B)
        def check(root, tmplate):
            if not root and tmplate:
                return False
            if not tmplate:
                return True
            if root.val != tmplate.val:
                return False
            return check(root.left, tmplate.left) and check(root.right, tmplate.right)
        return dfs(A, B)

1.4 226. 翻转二叉树

太巧妙了哇，想不到555

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return
        tmp = root.left
        root.left = self.invertTree(root.right)
        root.right = self.invertTree(tmp)
        return root

1.5 对称的二叉树

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
        def recur(L, R):
            if not L and not R:
                return True
            if not L or not R or L.val != R.val:
                return False
            return recur(L.left, R.right) and recur(L.right, R.left)
        return not root or recur(root.left, root.right)

1.6 从上到下打印二叉树

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrder(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        res, queue = [], collections.deque()
        queue.append(root)
        while queue:
            node = queue.popleft()
            res.append(node.val)
            if node.left: queue.append(node.left)
            if node.right: queue.append(node.right)
        return res

1.7 从上到下打印二叉树 II

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        res, queue = [], collections.deque()
        queue.append(root)
        while queue:
            tmp = []
            for i in range(len(queue)):
                node = queue.popleft()
                tmp.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            res.append(tmp)
        return res

1.8 从上到下打印二叉树 III

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        res, queue = [], collections.deque()
        queue.append(root)
        flag = False
        while queue:
            tmp = []
            for i in range(len(queue)):
                node = queue.popleft()
                tmp.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            if flag:
                tmp.reverse()
            flag = not flag
            res.append(tmp)
        return res

1.9 路径总和 II

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        res, path = [], []
        def recur(root, tar):
            if not root: return
            path.append(root.val)
            tar -= root.val
            if tar == 0 and not root.left and not root.right:
                res.append(list(path))
            recur(root.left, tar)
            recur(root.right, tar)
            path.pop()
        recur(root, targetSum)
        return res

1.10 二叉搜索树的第k大节点

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def kthLargest(self, root: TreeNode, k: int) -> int:
        path = []
        def dfs(root, path):
            if not root: return
            path.append(root.val)
            dfs(root.left, path)
            dfs(root.right, path)
            return path
        dfs(root, path)
        path.sort()
        path.reverse()
        return path[k-1]

1.11 二叉树的深度

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if not root:
            return 0
        res, queue = [], collections.deque()
        queue.append(root)
        while queue:
            tmp = []
            for i in range(len(queue)):
                node = queue.popleft()
                tmp.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            res.append(tmp)
        return len(res)

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if not root: return 0
        return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1
        

1.12 平衡二叉树

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isBalanced(self, root: Optional[TreeNode]) -> bool:
        if not root: return True
        return abs(self.deepth(root.left) - self.deepth(root.right)) <= 1 and \
            self.isBalanced(root.left) and self.isBalanced(root.right)
    def deepth(self, root):
            if not root: return 0
            return max(self.deepth(root.left), self.deepth(root.right)) + 1

1.13 求1+2+…+n

class Solution:
    def sumNums(self, n: int) -> int:
        if n == 1:
            return 1
        n += self.sumNums(n-1)
        return n

1.14 二叉搜索树的最近公共祖先

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        while root:
            if root.val < p.val and root.val < q.val:
                root = root.right
            elif root.val > p.val and root.val > q.val:
                root = root.left
            else:
                break
        return root

1.15 二叉树的最近公共祖先

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if not root or root == p or root == q:
            return root
        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)
        if not left and not right:
            return 
        if not left:
            return right
        if not right:
            return left
        return root

2 查找算法

2.1 数组中重复的数字

哈希表（Set）。

class Solution:
    def findRepeatNumber(self, nums: List[int]) -> int:
        dic = set()
        for num in nums:
            if num in dic:
                return num
            dic.add(num)
        return -1

2.2 二维数组中的查找

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        i, j = len(matrix) - 1, 0
        while i >= 0 and j < len(matrix[0]):
            if matrix[i][j] < target:
                j += 1
            elif matrix[i][j] > target:
                i -= 1
            else:
                return True
        return False

2.3 第一个只出现一次的字符

class Solution:
    def firstUniqChar(self, s: str) -> str:
        dict = {}
        for c in s:
            dict[c] = not c in dict
        for c in dict:
            if dict[c]:
                return c
        return ' '

2.4 在排序数组中查找数字 I

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        dict = collections.defaultdict(int)
        for c in nums:
            dict[c] += 1
        return dict[target]

2.5 0～n-1中缺失的数字

class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        i, j = 0, len(nums) - 1
        while i <= j:
            m = (i + j) // 2
            if nums[m] == m:
                i = m + 1
            else:
                j = m - 1
        return i

3 排序

3.1 最小的k个数

class Solution:
    def getLeastNumbers(self, arr: List[int], k: int) -> List[int]:
        arr.sort()
        return arr[:k]

3.2 扑克牌中的顺子

class Solution:
    def isStraight(self, nums: List[int]) -> bool:
        joker = 0
        nums.sort() # 数组排序
        for i in range(4):
            if nums[i] == 0: joker += 1 # 统计大小王数量
            elif nums[i] == nums[i + 1]: return False # 若有重复，提前返回 false
        return nums[4] - nums[joker] < 5 # 最大牌 - 最小牌 < 5 则可构成顺子

4 双指针

4.1 删除链表的节点

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def deleteNode(self, head: ListNode, val: int) -> ListNode:
        if head.val == val:
            return head.next
        pre, cur = head, head.next
        while cur and cur.val != val:
            pre, cur = cur, cur.next
        if cur:
            pre.next = cur.next
        return head

4.2 调整数组顺序使奇数位于偶数前面

class Solution:
    def exchange(self, nums: List[int]) -> List[int]:
        i, j = 0, len(nums) - 1
        while i < j:
            while i < j and nums[i] & 1 == 1: i += 1
            while i < j and nums[j] & 1 == 0: j -= 1
            nums[i], nums[j] = nums[j], nums[i]
        return nums

4.3 链表中倒数第k个节点

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
        former, latter = head, head
        for _ in range(k):
            former = former.next
        while former:
            former, latter = former.next, latter.next
        return latter

4.4 合并两个有序链表

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        head = ListNode()
        node = head
        while list1 and list2:
            if list1.val < list2.val:
                node.next, list1 = list1, list1.next
            else:
                node.next, list2 = list2, list2.next
            node = node.next
        node.next = list1 if list1 else list2
        return head.next

4.5 两个链表的第一个公共节点

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        A, B = headA, headB
        while A != B:
            A = A.next if A else headB
            B = B.next if B else headA
        return A

4.6 反转字符串中的单词

class Solution:
    def reverseWords(self, s: str) -> str:
        s = s.strip()         # 删除首尾空格
        strs = s.split()      # 分割字符串
        strs.reverse()        # 翻转单词列表
        return ' '.join(strs) # 拼接为字符串并返回

5 动态规划

5.1 斐波那契数列

class Solution:
    def fib(self, n: int) -> int:
        a, b = 0, 1
        for _ in range(n):
            a, b = b, (a + b) % 1000000007
        return a

5.2 青蛙跳台阶问题

class Solution:
    def numWays(self, n: int) -> int:
        a, b = 1, 1
        for _ in range(n):
            a, b = b, (a + b) % 1000000007
        return a

5.3 最大子数组和

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        for i in range(1, len(nums)):
            nums[i] += max(nums[i-1], 0)
        return max(nums)

5.4 把数字翻译成字符串

class Solution:
    def translateNum(self, num: int) -> int:
        s = str(num)
        a, b = 1, 1
        for i in range(2, len(s) + 1):
            a, b = (a + b if "10" <= s[i - 2:i] <= "25" else a), a
        return a

5.5 礼物的最大价值

class Solution:
    def maxValue(self, grid: List[List[int]]) -> int:
        row, column = len(grid), len(grid[0])
        for i in range(row):
            for j in range(column):
                left = grid[i][j - 1] if j - 1 >= 0 else 0
                upper = grid[i - 1][j] if i - 1 >= 0 else 0
                grid[i][j] += max(left, upper)
        return grid[row-1][column-1]

5.6 无重复字符的最长子串

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        res = tmp = i = 0
        for j in range(len(s)):
            i = j - 1
            while i >= 0 and s[i] != s[j]: i -= 1
            tmp = tmp + 1 if tmp < j - i else j - i
            res = max(res, tmp)
        return res
            

5.7 股票的最大利润

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        profit, cost = 0, float("+inf")
        for price in prices:
            cost = min(price, cost)
            profit = max(profit, price - cost)
        return profit

6 数据结构

6.1 替换空格

class Solution:
    def replaceSpace(self, s: str) -> str:
        s = s.split(" ")
        ss = ""
        for i in range(len(s)  - 1):
            ss += s[i] + "%20"
        ss += s[len(s) - 1]
        return ss

6.2 从尾到头打印链表

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reversePrint(self, head: ListNode) -> List[int]:
        stack = []
        while head:
            stack.append(head.val)
            head = head.next
        return stack[::-1]

6.3 用两个栈实现队列

class CQueue:

    def __init__(self):
        self.A, self.B = [], []

    def appendTail(self, value: int) -> None:
        self.A.append(value)

    def deleteHead(self) -> int:
        if self.B:
            return self.B.pop()
        if not self.A:
            return -1
        while self.A:
            self.B.append(self.A.pop())
        return self.B.pop()



# Your CQueue object will be instantiated and called as such:
# obj = CQueue()
# obj.appendTail(value)
# param_2 = obj.deleteHead()

6.4 反转链表

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        cur, pre = head, None
        while cur:
            tmp = cur.next
            cur.next = pre
            pre = cur
            cur = tmp
        return pre

6.5 左旋转字符串

class Solution:
    def reverseLeftWords(self, s: str, n: int) -> str:
        return s[n:] + s[:n]

7 分治算法

7.1 数值的整数次方

class Solution:
    def myPow(self, x: float, n: int) -> float:
        return x ** n

7.2 打印从 1 到最大的 n 位数

class Solution:
    def printNumbers(self, n: int) -> List[int]:
        res = []
        for i in range(1, 10 ** n):
            res.append(i)
        return res

8 位运算

8.1 191. 位1的个数

class Solution:
    def hammingWeight(self, n: int) -> int:
        x = list(bin(n))
        res = 0
        for i in x:
            if i == '1':
                res += 1
        return res

9 模拟

9.1 螺旋矩阵

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        if not matrix:
            return []
        l, r, t, b, res = 0, len(matrix[0]) - 1, 0, len(matrix) - 1, []
        while True:
            for i in range(l, r + 1):
                res.append(matrix[t][i])
            t += 1
            if t > b:
                break
            for i in range(t, b + 1):
                res.append(matrix[i][r])
            r -= 1
            if l > r:
                break
            for i in range(r, l - 1, -1):
                res.append(matrix[b][i])
            b -= 1
            if t > b:
                break
            for i in range(b, t - 1, -1):
                res.append(matrix[i][l])
            l += 1
            if l > r:
                break
        return res

9.2 验证栈序列

class Solution:
    def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool:
        stack, i = [], 0
        for num in pushed:
            stack.append(num)
            while stack and stack[-1] == popped[i]:
                stack.pop()
                i += 1
        return not stack