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通过索引的距离更改数组的值（c）

[英]Changing the values of array by the distance of the indexes (c)

原文 2018-02-16 13:30:27 7 1 c/ arrays/ binary

I'm having hard time with this one: I need to write a function in C that recieving a binary array and his size, and the function should calculate and replace the current values with the distance (by indexes) of each 1 to the closest 0. for example: if the function recieve that array {1,1,0,1,1,1,0,1} then the new values of the array should be {2,1,0,1,2,1,0,1} . 我很难用这个：我需要在C中编写一个接收二进制数组及其大小的函数，函数应该计算并用每个1到最近的距离（通过索引）替换当前值例如：如果函数接收到数组{1,1,0,1,1,1,0,1}则数组的新值应为{2,1,0,1,2,1,0,1} 。 It is known that the input has atleast 1 zero. 众所周知，输入至少为零。

So the first step I tought about was to locate pair of zeros (or just 1 if there is only 1) and set them as 2 indexes ( z1 , z2 ). 所以我想要的第一步是找到一对零（如果只有一个则只有1）并将它们设置为2个索引（ z1 ， z2 ）。 Then I set another index i that check everytime which zero is the closest to him (absolute value) and then the diffrence between i and z1 or z2 would be the new value. 然后我设置另一个索引i ，每次检查哪个零最接近他（绝对值），然后i和z1或z2之间的差异将是新值。

I have the plan but things are not going exactly as I planned. 我有计划，但事情并没有完全符合我的计划。 Basicly I deleted the code (it wasn't good anyway) so I would appreciate any help. 基本上我删除了代码（不管怎样都不好）所以我很感激任何帮助。 thanks! 谢谢！

1 个解决方案

This problem is based on two things 这个问题基于两件事

Keep an array left[i] which has the distance of nearest 0 from index i from left to right. 保持一个数组left[i] ，其距离索引i的距离最近为0 。
Keep an array right[i] which has the distance of nearest 0 from index i from right to left. 保持一个数组right[i] ，其距离索引i的距离最近为0 。

Both can be calculate in single loop iteration. 两者都可以在单循环迭代中计算。 O(n) . O(n) 。

Then for each position get the minimum value of left[i] and right[i] . 然后为每个位置获得left[i]和right[i]的最小值。 That will be the answer for 1 staying in position i . 这将是1留在i位置的答案。