Formatted question description: https://leetcode.ca/all/1089.html
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 <= arr.length <= 10000
0 <= arr[i] <= 9
Related Topics:
Array
Solution 1.
// OJ: https://leetcode.com/problems/duplicate-zeros/
// Time: O(N)
// Space: O(1)
class Solution {
public:
void duplicateZeros(vector<int>& A) {
int cnt = 0;
for (int n : A) cnt += n == 0;
if (cnt == 0) return;
for (int i = A.size() - 1; i >= 0; --i) {
cnt -= A[i] == 0;
int j = i + cnt;
if (j < A.size()) A[j] = A[i];
if (j + 1 < A.size() && A[i] == 0) A[j + 1] = 0;
}
}
};
Java
class Solution {
public void duplicateZeros(int[] arr) {
int length = arr.length;
int slow = 0, fast = 0;
while (fast < length) {
int num = arr[slow];
slow++;
if (num == 0)
fast += 2;
else
fast++;
}
if (fast == slow)
return;
if (arr[slow] == 0)
fast++;
while (fast >= length) {
int num = arr[slow];
slow--;
if (num == 0) {
fast -= 2;
if (fast == length - 2)
arr[length - 1] = 0;
} else
fast--;
}
while (fast > slow) {
int num = arr[slow];
slow--;
arr[fast] = num;
fast--;
if (num == 0) {
arr[fast] = num;
fast--;
}
}
}
}