双指针巧解复写零问题

复写零问题解析

问题描述
给定一个固定长度的数组，将所有出现的零复制一次，其余元素向右移动，超出数组长度的部分舍弃。例如：输入 [1,0,2,3,0,4,5,0]，输出 [1,0,0,2,3,0,0,4]。

双指针法原理

核心思想
通过快慢指针模拟元素复写过程：

快指针遍历原数组，标记需要复写的零。
慢指针指向新数组的当前位置，决定是否插入复写的零。

边界条件

数组末尾的零无需复写（因长度限制）。
复写零可能导致原数据被覆盖，需从后向前处理。

实现步骤

预处理：计算有效长度
遍历数组统计零的个数，确定最终有效元素的位置：

n = len(arr)
zeros = arr.count(0)
size = n + zeros

逆向填充
从有效长度的末尾开始填充数据，遇到零则复写：

i = n - 1  # 原数组指针
j = size - 1  # 新数组指针
while i >= 0 and j >= 0:
    if arr[i] != 0:
        if j < n:
            arr[j] = arr[i]
        j -= 1
    else:
        if j < n:
            arr[j] = 0
        j -= 1
        if j < n:
            arr[j] = 0
        j -= 1
    i -= 1

复杂度分析

时间复杂度：O(n)，仅需两次线性遍历。
空间复杂度：O(1)，原地修改数组，无额外空间消耗。

示例代码

完整Python实现：

def duplicateZeros(arr):
    n = len(arr)
    zeros = arr.count(0)
    size = n + zeros
    i, j = n - 1, size - 1
    
    while i >= 0 and j >= 0:
        if arr[i] != 0:
            if j < n:
                arr[j] = arr[i]
            j -= 1
        else:
            if j < n:
                arr[j] = 0
            j -= 1
            if j < n:
                arr[j] = 0
            j -= 1
        i -= 1

关键点总结

逆向操作：避免正向复写导致数据覆盖丢失。
双指针协作：快指针定位数据，慢指针控制写入位置。
边界检查：确保写入时不越界，舍弃超出部分。

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