二分查找遞歸與非遞歸總結

二分查找雖然簡單，但是很基礎，筆試以及面試的命中率很高，因此再次複習，並將易混淆之處指出，以免犯錯。

參考：http://blog.163.com/hjzjoy@126/blog/static/5375838320071019478271/

二分查找的非遞歸查找，採用while(low<=high)來控制查找的進行。

int BiSearch(int A[], int low, int high, int x)
{
    while(low <= high)
    {
              int mid = (low + high)/2;
              if(x > A[mid])
                   low = mid + 1;
              else if(x < A[mid])
                   high = mid - 1;
              else
                  return mid;
    }
    return -1;
}

遞歸實現

第一個遞歸是我自己實現的版本，結構和非遞歸差不多，但是需要在結尾加一個判斷條件，將需要的下標返回。

int BiSearchNew1(int A[], int low, int high, int x)
{
    int mid;
	mid = (low + high)/2;
	if(low <= high)
	{
		if(x > A[mid])
		{
			mid = BiSearchNew1(A, mid + 1, high, x);
		}
		else if(x < A[mid])
		{
			mid = BiSearchNew1(A, low, mid - 1, x);
		}
		else
		    return mid;
		if(mid >= 0)
		{
			return mid;
		}
	}
	return -1;
}

通過查閱，發現有更加方便的辦法，附於下：

int BiSearchNew2(int A[], int low, int high, int x)
{
    int mid;
    if(low > high)
           return -1;
    mid = (low + high)/2;
    if(x == A[mid])
         return mid;
    
    if(x < A[mid])
         return BiSearchNew2(A, low, mid - 1, x);
    else
        return BiSearchNew2(A, mid + 1, high, x);
}

下面是測試程序：

int main()
{
    int A[7] = {1, 3, 4, 6, 7, 9, 13};
    int i;
    int j;
    for(j = 0; j < 7; j++)
    {
          i = BiSearch(A, 0, 6, A[j]);
          printf("%d ", i);
    }
    printf("\n");
    for(j = 0; j < 7; j++)
    {
          i = BiSearchNew1(A, 0, 6, A[j]);
          printf("%d ", i);
    }
    printf("\n");
    for(j = 0; j < 7; j++)
    {
          i = BiSearchNew2(A, 0, 6, A[j]);
          printf("%d ", i);
    }
    printf("\n");
    getch();
    return 0;
}