Time complexity of the code below?

Can someone tell me the time complexity for the following code?

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
char a[100]= "G开发者_JS百科osh I am confused :D";
int i,count= -1,display_ToVal= strlen(a)-1, display_FromVal;

for( i=strlen(a)-1 ; i>=0 ; i=i+count)
{
        if ( (a[i] == ' ' || i == 0) && count == -1)
        {
         cout << " ";
         display_FromVal = i;
         count = 1;
         if ( i == 0 )
                cout << a[i];
         continue;
        }       

        else if( count == 1 && i == display_ToVal)
        {
         cout << a[i];
         display_ToVal = display_FromVal - 1;
         i = display_FromVal;
         count = -1;
         if(display_FromVal == 0)
                 break;
         else
                 continue;
        }

        else if (count == 1)
         cout << a[i];

        else
         continue;
}

return 1;
} 

I am really confused as to whether this can be classified as O(n). Please help, thank you in advance.

---

The algorithm can be summarrized in pseudo-code as :

1. mark the current position
2. go backward one character at a time until a space is found or end of input is reached
3. now go forward copying each character to output, then go back to 1., except if eoi is reached

So the input is traversed once in reverse, and one more time forward, but without coming back to a previously read position in either step 2. or 3. And when switching from step 3. to 1. it directly adjust the iterator. The count variable is used to track the state of the algorithm (it is in fact a simple state-machine). It is also reused to define the direction of the iteration.

So, the algorithm is in fact O(n).

---

For more clarity, it could be rewritten as this, without changing the complexity:

void printStringWithWordReversed(const char* a) {
    int i,j,display_ToVal= strlen(a)-1, display_FromVal;
    for( i=display_ToVal; i>=0 ; i=i+-1)
    {
        if ( (a[i] == ' ' || i == 0))
        {
         // When entering this branch, we are switching from state 2 to
         // state 3 (this is the content of the first branch).
         cout << " ";
         display_FromVal = i;
         if ( i == 0 )
                cout << a[i];
         // This loop correspond to the state 3, and is equivalent to the
         // previous code in the particular case when count == 1.
         for (j = display_FromVal+1; j <= display_ToVal; j=j+1)
         {
             cout << a[j];
         }
         // This postlude correspond to the transition from state 3 to state 1
         // and correspond to the second branch in the original algorithm.
         display_ToVal = display_FromVal - 1;
         if ( i == 0 )
            break;
         continue;
        }       
    }
}

So we lookup each word starting from the end and output them in correct order. This is clearly O(n) with both implementation (both in time and space if we assume that cout insertion operator for char is O(1)) since adding a fixed number (here two) of O(n) algorithm is still O(n) (the constant is ignored).

---

"{for( i=strlen(a)-1 ; i>=0 ; i=i+count)}"

There is only one for loop in your code and its index i is varying linearly . Thats why its O(n)