题目如下:
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
解题思路:从头开始遍历arr,如果arr[i]等于0,在i+1位置插入0,索引移动到i+2;如果不为0,索引移动到i+1。
代码如下:
class Solution(object): def duplicateZeros(self, arr): """ :type arr: List[int] :rtype: None Do not return anything, modify arr in-place instead. """ length = len(arr) inx = 0 while inx < length: if arr[inx] != 0: inx += 1 continue else: arr.insert(inx,0) arr.pop(-1) inx += 2
原文地址:https://www.cnblogs.com/seyjs/p/11044681.html
时间: 2024-10-12 17:32:31