1089. Duplicate Zeros - Easy

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

Hint: Iterate through the array backwards. You know whether an integer should be written or not based on how many zeroes are remaining in the array.

 

注意在第二次iterate覆盖赋值的时候，超出原数组长度的部分不用管，只在原数组长度范围内写入新值就行

time: O(n), space: O(1)

class Solution {
    public void duplicateZeros(int[] arr) {
        int cnt = 0;    // count zeros
        for(int n : arr) {
            if(n == 0) {
                cnt++;
            }
        }
        
        int n = arr.length + cnt;
        int i = n - 1, j = arr.length - 1;
        while(i >= 0 && j >= 0) {
            if(arr[j] == 0) {
                if(i < arr.length) {
                    arr[i] = 0;
                }
                i--;
                if(i < arr.length) {
                    arr[i] = 0;
                }
            } else {
                if(i < arr.length) {
                    arr[i] = arr[j];
                }
            }
            i--;
            j--;
        }
    }
}