剑指offer刷题记录｜顺序刷题第二周

剑指offer 16 数值的整数次方

class Solution:
    def myPow(self, x: float, n: int) -> float:
        # # 法一：超时
        # res = 1
        # if n > 0:
        #     for i in range(n):
        #         res *= x
        # # 考虑n<0
        # if n < 0:
        #     for i in range(abs(n)):
        #         res /= abs(x)
        # return res

        # 递归
        def cal_result(x,n):
            if n==1: return x
            # 递归哦
            num = cal_result(x, n//2)
            if n % 2 == 0:
                return num * num
            else:
                return num * num * x

        if n==0: return 1
        # 考虑n为负，则倒数
        if n<0:
            return 1/cal_result(x,-n)
        return cal_result(x,n)

剑指offer18 删除链表某个节点

好难 有点想不明白

class Solution:
    def deleteNode(self, head: ListNode, val: int) -> ListNode:
      	# 第一个val就是的时候，直接返回后面部分
        if head.val == val: return head.next
        pre = head
        while pre.next and pre.next.val!=val:
            pre = pre.next
        if pre.next:
            pre.next = pre.next.next
        return head # not pre

剑指offer17 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

剑指offer 15 二进制中1的个数

class Solution:
    def hammingWeight(self, n: int) -> int:
        # 法一：使用语言特性，不过输入的n是整数就不是二进制？？？
        cnt=Counter(str(bin(n)))
        return cnt['1']

        # 法二：使用语言特性
        return bin(n).count('1')

        # 法三：按位与运算
        res = 0
        while n:
            res += n&1
            n >>= 1
        return res

剑指offer14-II 剪绳子

class Solution:
    def cuttingRope(self, n: int) -> int:
        if n <= 3 : return n-1
        a, b = n//3, n%3
        # math.pow() 时间复杂度低，但是有溢出
        if b == 0: res = int(pow(3, a))
        if b == 1: res = int(pow(3, a-1) * 4) 
        if b == 2: res = int(pow(3, a) * 2)
        return res % 1000000007

剑指offer 14-I 剪绳子

class Solution:
    def cuttingRope(self, n: int) -> int:
        # 长度为n的绳子分m段 长度乘积最大
        # 每段长度相同时 是最佳的； 分成三段 是最佳的
        # 所以每段长度3 余数0则3的n次方 余1则有一段是4 余2则有一段是2
        a, b = n//3, n%3
        if n<=3 : return n-1
        if b==0: return int(math.pow(3,a))
        if b==1: return int(math.pow(3, a-1)*4) 
        if b==2: return int(math.pow(3, a)*2)

剑指offer 13 机器人的运动范围

class Solution:
    def movingCount(self, m: int, n: int, k: int) -> int:
        def digital(x):
            sumres = 0
            while x:
                sumres += x%10
                x //= 10
            return sumres

        visited = set([(0,0)])
        for i in range(m):
            for j in range(n):
                # 只要考虑上一个是从左边过来还是上面过来就好了
                if digital(i)+digital(j)<=k and ((i-1,j) in visited or (i,j-1) in visited):
                    visited.add((i,j))
        return len(visited)

剑指offer 12 矩阵中的路径

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:

        def DFS(i, j, seen):
            # 停止条件
            if len(seen)==len(word):
                return True
            # 遍历周围四个区域，没有False，有则继续
            for (x,y) in [(i-1,j), (i+1,j), (i,j-1), (i,j+1)]:
                if 0<=x<row and 0<=y<column and (x,y) not in seen and board[x][y]==word[len(seen)]:
                    seen.add((x,y))
                    if DFS(x,y,seen):
                        return True
                    seen.remove((x,y))
            return False

        if word == []: return True
        row, column = len(board), len(board[0])
        for i in range(row):
            for j in range(column):
                # DFS + 剪枝
                if board[i][j]==word[0] and DFS(i, j, {
     (i,j)}):
                    return True
        return False

剑指offer 11

class Solution:
    def minArray(self, numbers: List[int]) -> int:
        # 法一：递增嘛，遇到gap就是答案了
        for i in range(len(numbers)-1):
            if numbers[i]>numbers[i+1]:
                return numbers[i+1]
        return min(numbers[0],numbers[-1])

        # 法二：直接返回最小值呗
        return min(numbers)

剑指offer 10-2

class Solution:
    def numWays(self, n: int) -> int:
        if n == 0: return 1
        if n == 1: return 1
        # 1, 1, 2, 3, 5, 8, 13, 21 
        # f(n) = f(n-1) + f(n-2) 
        # 考虑是从前一个跳过来的，还是前两个跳过来的
        # 从两个之前过来不能一步一步过来，因为那就和从前一个过来重复了
        a, b = 1, 1 
        for i in range(2, n+1):
            a, b = b, a+b
        return b % 1000000007