LeetCode1089 replication zero

problem:

Give you a fixed-length array of integers  arr, each zero you the array are replication occurs again, and the remaining elements of pan right.

Note: Please do not write element in position over the length of the array.

Requirements: Please enter the array of the above modifications on the spot, do not return anything from a function.

 

Example 1:

Input: [1,0,2,3,0,4,5,0] 
Output: null 
explained: after the function call, the array input will be modified to: [1,0,0,2,3,0,0 , 4]

Example 2:

Input: [1,2,3] 
Output: null 
explained: after the function call, the array input will be modified to: [1,2,3]

 

prompt:

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

Links: https://leetcode-cn.com/contest/weekly-contest-141/problems/duplicate-zeros/

analysis:

Do not write more than is required in the position of the element array length, you can traverse again, experiencing zero will erase the last one, and then insert a 0 at the current position.

 

AC Code：

 

 1 class Solution {
 2 public:
 3     void duplicateZeros(vector<int>& arr) {
 4         int length = arr.size();
 5         for (int i = 0; i < length; i++)
 6         {
 7             if (arr[i] == 0)
 8             {
 9                 arr.erase(arr.end()-1);
10                 arr.insert(arr.begin()+i, 0);
11                 i++;
12             }
13         }        
14     }
15 };

 

other:

1. The first code

 1     class Solution {
 2         public void duplicateZeros(int[] arr) {
 3             int n = arr.length;
 4             int[] a = Arrays.copyOf(arr, n);
 5             int p = 0;
 6             for(int i = 0;i < n;i++){
 7                 if(a[i] == 0){
 8                     if(p < n)arr[p++] = 0;
 9                     if(p < n)arr[p++] = 0;
10                 }else{
11                     if(p < n)arr[p++] = a[i];
12                 }
13             }
14         }
15     }    

Construction backup input data, and modify the original position.

2, vector remove elements erase,

vector<int> vec;

vec.erase (vec.begin () + index); // delete vec [index] Data

vec.insert (vec.begin () + index, value); // value is inserted in position index.