Question

给定数组

$array = [ 0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0];

对所有0进行分组并将它们表示为起始和结束索引的最快算法是什么？

（即）输出应为（0,5），（10,17）......

Answer 1

伪代码将是这样的 -

for(int i = 0; i < array.length; i++) {
     if(array[i] == 0) {
         int left = i;
         while(i + 1 < array.length && array[i + 1] == 0) {
              i++; 
         }
         // print range [left, i] 
     }    
}

时间复杂度为O(n)，其中n为数组长度。空间复杂性是不变的。

Answer 2

让我以伪代码形式编写多线程方法：

Function ZeroHunter()
  Split the array into N segments (where N is multiple of number of threads available)
  Create a Dictionary<SegmentNumber, Result>. Let's call it DicResults.

  For i = 0 to Number of segments
    Result = ProcessSegmentOnNewThread(S)
    DicResults.Add(i, Result)
  Next

  Wait for all threads to complete

  ConcatResults(DicResults)
End Function

ProcessSegmentOnNewThread的代码：

Function ProcessSegmentOnNewThread(Segment S)
  Create a new thread

  List<int, int> lst;
  while i < S.Length
    if S[i] == 0 Then 
     NewEntry = lst.Add(i, i)
     while i < S.Length and S[i] == 0
       NewEntry.End++;
       i++
     end while
    end if

    i++
  end while

  Return the results in the form of List<int, int>
End Function

ConcatResults的代码（）：

Function ConcatResults(DicResults)
  Sort the input dictionary on segment number (the dictionary key that is)

  For i = 0 to DicResults.Count - 1
    If DicResult[i].End = DicResult[i+1].Start - 1
      Set DicResult[i].End = DicResult[i+1].End
      Remove DicResult[i+1] from the dictionary
    End If
  Next

  Return DicResult
End Function

需要注意的是，虽然这个算法会击败非常大的数组的非线程算法，但你不应该将它用于较小的数组，因为创建线程和连接结果的开销将超过线程的潜在好处。

Answer 3

试试这个好友...... !!希望它的罚款：D

#include <stdio.h>

int main()
{
int a[12] = {0,0,0,0,1,1,1,1,0,0,0,0};
int i;
int t = 0;
  for( i = 0; i < (int)( sizeof(a) / sizeof(a[0])); i++) {
     if(a[i] == t) {
         int k = i;
         while(i + 1 < (int)( sizeof(a) / sizeof(a[0])) && a[i + 1] == 0) {
              i++; 
         }
         printf("range(%d,%d)\n",k,i);
     }    
}
}

查找数组中数据范围的算法

3 个答案: