[LeetCode - Python]剑指 Offer 21. 调整数组顺序使奇数位于偶数前面(Easy)；27. 移除元素(Easy)；925. 长按键入（Easy 我感觉是Medium）

1. 题目

剑指 Offer 21. 调整数组顺序使奇数位于偶数前面(Easy);

1.代码：

class Solution:
    def exchange(self, nums: List[int]) -> List[int]:
        # 两个List存
        Ji = []
        Ou = []
        for i in nums :
            if i%2 != 0 :
                Ji.append(i)
            else :
                Ou.append(i)
        return Ji + Ou

class Solution:
    def exchange(self, nums: List[int]) -> List[int]:
        # 对撞双指针
        ret = [0 for _ in nums]
        left,right = 0 , len(nums)-1
        for i in range(len(nums)):
            if nums[i]%2 !=0 :
                ret[left] = nums[i]
                left+=1
            else :
                ret[right] = nums[i]
                right-=1
        return ret

class Solution:
    def exchange(self, nums: List[int]) -> List[int]:
        # 快慢指针
        # 慢指针存最后一个奇数位置，快指针搜索下一个奇数位置
        fast,slow = 0 , 0
        while fast < len(nums):
            if nums[fast]%2 != 0 :
                nums[slow] , nums[fast] = nums[fast] ,nums[slow]
                slow+=1
            fast+=1
        return nums

2.题目

27. 移除元素(Easy)

2.代码：

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        # 倒序快慢指针
        fast , low = len(nums) -1, len(nums)-1
        while fast >= 0 :
            if nums[fast] == val:
                nums[low],nums[fast] =nums[fast],nums[low]
                low-=1
            fast-=1
        return low+1

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        # 快慢指针
        fast , slow = 0 , 0
        while fast < len(nums):
            if nums[fast]!= val:
                nums[slow],nums[fast]=nums[fast],nums[slow]
                slow+=1
            fast+=1
        return slow

3.题目

925. 长按键入（Easy）

3.代码：

class Solution:
    def isLongPressedName(self, name: str, typed: str) -> bool:
        # 双指针
        # name → N1
        # typed →N2
        N1,N2 = 0,0
        while N1<len(name) and N2<len(typed) :
            if name[N1]==typed[N2]:
                N1+=1
                N2+=1
            elif N2>0 and typed[N2] == typed[N2-1]:
                N2+=1
            else :
                return False
        while N2<len(typed) and typed[N2]==typed[N2-1]:
            N2+=1
        if N1 == len(name) and N2 == len(typed):
            return True
        else:
            return False
            

class Solution:
    def isLongPressedName(self, name: str, typed: str) -> bool:
        # 双指针 -- 看完题解简化版本
        # name → N1
        # typed →N2
        N1 ,  N2 = 0 , 0
        while N2 < len(typed):          # 这里只考虑N2 是因为若返回True 则必须 N2>=N1
            if N1<len(name) and name[N1]==typed[N2] : # 防止N1溢出，当二者相等,均+1
                N1 += 1
                N2 += 1
            elif N2>0 and typed[N2]==typed[N2-1]:   # 考虑跟上一个相等的时候，要考虑List仅有一个数时候[0] ==[0-1]
                N2 +=1
            else :
                return False
        return N1 == len(name)  #防止出现 len(name) > len(typed)出现