README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1000-1099/1089.Duplicate%20Zeros/README_EN.md	1262	Weekly Contest 141 Q1	Array Two Pointers

1089. Duplicate Zeros

中文文档

Description

Given a fixed-length integer array arr, 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 and do not return anything.

 

Example 1:

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

Example 2:

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

 

Constraints:

• 1 <= arr.length <= 104
• 0 <= arr[i] <= 9

Solutions

Solution 1

Python3

class Solution:
    def duplicateZeros(self, arr: List[int]) -> None:
        """
        Do not return anything, modify arr in-place instead.
        """
        n = len(arr)
        i, k = -1, 0
        while k < n:
            i += 1
            k += 1 if arr[i] else 2
        j = n - 1
        if k == n + 1:
            arr[j] = 0
            i, j = i - 1, j - 1
        while ~j:
            if arr[i] == 0:
                arr[j] = arr[j - 1] = arr[i]
                j -= 1
            else:
                arr[j] = arr[i]
            i, j = i - 1, j - 1

Java

class Solution {
    public void duplicateZeros(int[] arr) {
        int n = arr.length;
        int i = -1, k = 0;
        while (k < n) {
            ++i;
            k += arr[i] > 0 ? 1 : 2;
        }
        int j = n - 1;
        if (k == n + 1) {
            arr[j--] = 0;
            --i;
        }
        while (j >= 0) {
            arr[j] = arr[i];
            if (arr[i] == 0) {
                arr[--j] = arr[i];
            }
            --i;
            --j;
        }
    }
}

C++

class Solution {
public:
    void duplicateZeros(vector<int>& arr) {
        int n = arr.size();
        int i = -1, k = 0;
        while (k < n) {
            ++i;
            k += arr[i] ? 1 : 2;
        }
        int j = n - 1;
        if (k == n + 1) {
            arr[j--] = 0;
            --i;
        }
        while (~j) {
            arr[j] = arr[i];
            if (arr[i] == 0) arr[--j] = arr[i];
            --i;
            --j;
        }
    }
};

Go

func duplicateZeros(arr []int) {
	n := len(arr)
	i, k := -1, 0
	for k < n {
		i, k = i+1, k+1
		if arr[i] == 0 {
			k++
		}
	}
	j := n - 1
	if k == n+1 {
		arr[j] = 0
		i, j = i-1, j-1
	}
	for j >= 0 {
		arr[j] = arr[i]
		if arr[i] == 0 {
			j--
			arr[j] = arr[i]
		}
		i, j = i-1, j-1
	}
}

Rust

impl Solution {
    pub fn duplicate_zeros(arr: &mut Vec<i32>) {
        let n = arr.len();
        let mut i = 0;
        let mut j = 0;
        while j < n {
            if arr[i] == 0 {
                j += 1;
            }
            j += 1;
            i += 1;
        }
        while i > 0 {
            if arr[i - 1] == 0 {
                if j <= n {
                    arr[j - 1] = arr[i - 1];
                }
                j -= 1;
            }
            arr[j - 1] = arr[i - 1];
            i -= 1;
            j -= 1;
        }
    }
}

C

void duplicateZeros(int* arr, int arrSize) {
    int i = 0;
    int j = 0;
    while (j < arrSize) {
        if (arr[i] == 0) {
            j++;
        }
        i++;
        j++;
    }
    i--;
    j--;
    while (i >= 0) {
        if (arr[i] == 0) {
            if (j < arrSize) {
                arr[j] = arr[i];
            }
            j--;
        }
        arr[j] = arr[i];
        i--;
        j--;
    }
}