数组比较问题

Question

有两个排好序的数组,如何比较两个数组相同的元素个数,最好不使用两个循环.

Answer 1

for(int cnt;cnt < strA.Length && cnt < strB.Length;cnt++)
{
   if(strA[cnt] != strB[cnt])
      MessageBox.Show("元素不一样！");
}是这个意思么？比较元素？

Answer 2

假设2个数组
A B简称爬杆法
把B当成杆，A的元素当成爬杆者
从A中拿最小的元素和B中从小到大元素比较
  如果查找到B中有比它大且不相当的元素，
  则从A中换一个元素接着爬

Answer 3

int i;
for(int cnt;cnt < strA.Length && cnt < strB.Length;cnt++) 
{ 
   if(strA[cnt] ！= strB[cnt]) 
      MessageBox.Show("元素不一样！"); 
   else
      i++;
}
MessageBox.Show("strA于strB中相同元素的个数是：" + i.ToString());  z这样差不多了吧。

Answer 4

            int cnt = 0;
            string[] strA = new string[10];
            string[] strB = new string[10];
            for (int j = 0; j < strA.Length; j++)
            {
                for (int i = 0; i < strB.Length; i++)
                {
                    if (strA[j] == strB[i])
                        cnt++;
                }
            }
            MessageBox.Show("相同元素个数：" + cnt.ToString());
这次对了，绕

Answer 5

zhchg6666 所说的爬杆法好像也要做两次循环,有没有不用循环两次的方法?数组是排好序的,有没有更好的方法

Answer 6

最好不使用两个循环.
for和if是运行最快的了，多套几个不会影响速度的，而且简单，何必自找烦恼呢？

Answer 7

    /// <re>
    ///  arrA and arrB should be already sorted
    /// </re>
    static void PrintSameElements(int[] arrA, int[] arrB)
    {
        int floor = 0;
        foreach (int b in arrB)
        {
            for (int i = floor; i < arrA.Length; i++)
            {
                if (b == arrA[i]) { Console.WriteLine(b); break; }
                else if (b < arrA[i]) { floor = i; break; }
            }
        }
    }

Answer 8

int EquallCount=0;
int i,j=0;do
{
   if(a[i]>b[j])
   {
     j++;
   }
   else if(a[i]<b[j])
   {
     i++;
   }
   else
   {
     count++
   }
}while(i!=a.count||j!=b.count)return count;//楼主是这个意思吗？

Answer 9

This one is better :-)

Answer 10

更改了一些错误（原代码见10楼）：        static int f(int[] a, int[] b)
        {
            int count = 0;
            int i = 0, j = 0;            while (i < a.Length && j < b.Length)
            {
                if (a[i] > b[j])
                {
                    j++;
                }
                else if (a[i] < b[j])
                {
                    i++;
                }
                else
                {
                    i++;
                    j++;
                    count++;
                }
            }            return count;
        }

Answer 11

一个数组中元素可以重复吗？这样的话算不算重复count增加吗/int[] a = new int[] { 1, 3, 4, 5, 6, 6, 7 };
int[] b = new int[] { 3, 3, 4, 6, 7, 9 };

Answer 12

        static int compareTowArray(int[] nAAry, int[] nBAry)
        {
            int nCount = 0;
            int i = 0, j = 0;
            Boolean bASolved = false, bBSolved = false;
            while (!bASolved || !bBSolved)
            {
                if ((i < nAAry.Length) && (j < nBAry.Length))
                {
                    if (nAAry[i] <= nBAry[j])
                    {
                        if (nAAry[i] == nBAry[j])
                        {
                            nCount++;
                        }
                        i++;
                    }
                    else
                    {
                        j++;
                    }
                }
                else
                {
                    if (i == nAAry.Length)
                    {
                        i--;
                        j++;
                        bASolved = true;
                        continue;
                    }                    if (j == nBAry.Length)
                    {
                        j--;
                        i++;
                        bBSolved = true;
                        continue;
                    }
                }
            }            return nCount;
        }

Answer 13

并不简单啊，修改版        static int compareTowArray(int[] nAAry, int[] nBAry)
        {
            int nCount = 0;
            int i = 0, j = 0;
            Boolean bASolved = false, bBSolved = false;
            int nRecordi = -2, nRecordj = -2;
            while (!bASolved || !bBSolved)
            {
                if ((i < nAAry.Length) && (j < nBAry.Length))
                {
                    if (nAAry[i] <= nBAry[j])
                    {
                        if (nAAry[i] == nBAry[j])
                        {
                            if (!(nRecordi + 1 == i && nRecordj == j) && !(nRecordj + 1 == j && nRecordi == i))
                            {
                                nCount++;
                                nRecordi = i;
                                nRecordj = j;
                            }
                        }
                        i++;
                    }
                    else
                    {
                        j++;
                    }
                }
                else
                {
                    if (i == nAAry.Length)
                    {
                        i--;
                        j++;
                        bASolved = true;
                        continue;
                    }                    if (j == nBAry.Length)
                    {
                        j--;
                        i++;
                        bBSolved = true;
                        continue;
                    }
                }
            }            return nCount;
        }