[LeetCode] 1089. Duplicate Zeros

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Description

Given a fixed length array arr of integers, duplicate each occurrence of zero, shifting the remaining elements to the right.

Note that elements beyond the length of the original array are not written.

Do the above modifications to the input array in place, do not return anything from your function.

Example 1:

Input: [1,0,2,3,0,4,5,0]
Output: null
Explanation: After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]

Example 2:

Input: [1,2,3]
Output: null
Explanation: After calling your function, the input array is modified to: [1,2,3]

Note:

1. 1 <= arr.length <= 10000
2. 0 <= arr[i] <= 9

 

思路

题意：给定一个固定长度的整数数组arr，复制每次出现的零，将剩余的元素向右移动。不使用其他辅助数据结构对输入数组进行上述修改。

题解：不使用其他辅助数据结构对数组进行原地修改，那么通过一个变量来记录数组中零值的个数，从后往前遍历，交换数组中的值。

 

+ View Code?

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static const auto io_sync_off = []() {     // turn off sync     std::ios::sync_with_stdio(false);     // untie in/out streams     std::cin.tie(nullptr);     return nullptr; }();   class Solution { public:     void duplicateZeros(vector<int>& arr) {         int zeroCnt = 0;         int size = arr.size();         for (int i = 0; i < size; i++){             if (arr[i] == 0)    zeroCnt++;         }         int j = size + zeroCnt;         for (int i = size - 1; i >= 0; i--){             if (--j < size)    arr[j] = arr[i];             if (arr[i] == 0 && --j < size){                 arr[j] = 0;             }         }     } };