复写零问题描述

给定一个固定长度的整数数组 arr，将其中所有的零元素复写（即每个零后面插入一个零），并保持数组长度不变。超过数组长度的元素被丢弃。

示例：输入：[1,0,2,3,0,4,5,0]
输出：[1,0,0,2,3,0,0,4]

双指针解法核心思路

使用双指针模拟复写过程，分为两步：

遍历模拟：通过快慢指针确定复写零后的有效元素位置。
逆向填充：从后向前填充数组，避免覆盖未处理的元素。

具体实现步骤

第一步：定位有效元素位置

初始化快指针 fast 和慢指针 slow，均从 0 开始。
遍历数组，当遇到零时，fast 多走一步（模拟复写）。
当 fast 达到或超过数组末尾时停止。

fast, slow = 0, 0
n = len(arr)
while fast < n:
    if arr[slow] == 0:
        fast += 1
    slow += 1
    fast += 1

第二步：逆向填充数组

从 slow-1 开始逆向遍历，将元素依次放入数组末尾。
遇到零时，复写两次（注意边界条件）。

slow -= 1  # 回退到最后有效位置
fast = n - 1
while fast >= 0:
    if arr[slow] != 0:
        arr[fast] = arr[slow]
    else:
        arr[fast] = 0
        if fast - 1 >= 0:
            arr[fast - 1] = 0
            fast -= 1
    slow -= 1
    fast -= 1

边界条件处理

末尾零处理：当最后一个有效元素是零且可能越界时，仅复写一次。
索引校验：逆向填充时检查 fast-1 是否合法。

复杂度分析

时间复杂度：O(n)，两次线性遍历。
空间复杂度：O(1)，原地修改数组。

完整代码示例

def duplicateZeros(arr):
    n = len(arr)
    fast, slow = 0, 0
    while fast < n:
        if arr[slow] == 0:
            fast += 1
        slow += 1
        fast += 1
    
    slow -= 1
    fast = n - 1
    while fast >= 0:
        if arr[slow] != 0:
            arr[fast] = arr[slow]
        else:
            arr[fast] = 0
            if fast - 1 >= 0:
                arr[fast - 1] = 0
                fast -= 1
        slow -= 1
        fast -= 1

关键点总结

快慢指针模拟复写后的逻辑长度。
逆向填充避免数据覆盖，确保稳定性。
边界条件需仔细处理，特别是末尾零的复写。

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#牛客AI配图神器#

双指针高效复写零技巧

复写零问题描述

双指针解法核心思路

具体实现步骤

边界条件处理

复杂度分析

完整代码示例

关键点总结

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