剑指offer 03. 数组中重复的数字 04. 二维数组中的查找 05. 替换空格 06. 从尾到头打印链表

面试题03. 数组中重复的数字
双循环果然又超时，直接用sort

class Solution:
    def findRepeatNumber(self, nums: List[int]) -> int:
        nums.sort()
        for i in range(1,len(nums)):
            if nums[i]==nums[i-1]:
                return nums[i]

最快答案是用字典

class Solution:
    def findRepeatNumber(self, nums: List[int]) -> int:
        repeatDict = {}
        for num in nums:
            if num not in repeatDict:
                repeatDict[num] = 1
            else:
                return num

面试题04. 二维数组中的查找
左下角标志数法

class Solution:
    def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
        i,j=len(matrix)-1,0
        while i>=0 and j<= len(matrix[0])-1:
            if matrix[i][j]<target:
                j+=1
            elif matrix[i][j]>target:
                i-=1
            else:
                return True
        return False

面试题05. 替换空格
直接用replace或者先split再join

class Solution:
    def replaceSpace(self, s: str) -> str:
        #s=s.replace(' ','%20')
        s='%20'.join(s.split(' '))
        return s

也可以开个新数组

class Solution:
    def replaceSpace(self, s: str) -> str:
        #s=s.replace(' ','%20')
        #s='%20'.join(s.split(' '))
        news=[]
        for i in s:
            if i==' ':
                news.append('%20')
            else:
                news.append(i)
        return ''.join(news)

面试题06. 从尾到头打印链表

• 反转

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

题解里的两行递归：

class Solution:
    def reversePrint(self, head: ListNode) -> List[int]:
        if not head: return []
        return self.reversePrint(head.next) + [head.val]

作者：z1m
链接：https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/solution/python3-c-by-z1m/
来源：力扣（LeetCode）
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