实现一个高效的算法来找到两个字符串的交集

Question

Implement an algorithm that takes two strings as input, and returns the intersection of the two, with each letter represented at most once.

Algo: (considering language used will be c#)

1. Convert both strings into char array
2. take the smaller array and generate a hash table for it with key as the character and value 0
3. Now Loop through the other array and increment the count in hash table if that char is present in it.
4. Now take out all char for hash table whose value is > 0.
5. These are intersection values.

This is an O(n), solution but is uses extra space, 2 char arrays and a hash table

Can you guys think of better solution than this?

Answer 1

How about this ...

var s1 = "aabbccccddd";
var s2 = "aabc";

var ans = s1.Intersect(s2);

Answer 2

Haven't tested this, but here's my thought:

1. Quicksort both strings in place, so you have an ordered sequence of characters
2. Keeping an index into both strings, compare the "next" character from each string, pick and output the first one, incrementing the index for that string.
3. Continue until you get to the end of one of the strings, then just pull unique values from the rest of the remaining string.

Won't use additional memory, only needs the two original strings, two integers, and an output string (or StringBuilder). As an added bonus, the output values will be sorted too!

Part 2: This is what I'd write (sorry about the comments, new to stackoverflow):

private static string intersect(string left, string right)
{
  StringBuilder theResult = new StringBuilder();

  string sortedLeft = Program.sort(left);
  string sortedRight = Program.sort(right);

  int leftIndex = 0;
  int rightIndex = 0;

  //  Work though the string with the "first last character".
  if (sortedLeft[sortedLeft.Length - 1] > sortedRight[sortedRight.Length - 1])
  {
    string temp = sortedLeft;
    sortedLeft = sortedRight;
    sortedRight = temp;
  }

  char lastChar = default(char);
  while (leftIndex < sortedLeft.Length)
  {
    char nextChar = (sortedLeft[leftIndex] <= sortedRight[rightIndex]) ? sortedLeft[leftIndex++] : sortedRight[rightIndex++];

    if (lastChar == nextChar) continue;

    theResult.Append(nextChar);
    lastChar = nextChar;
  }

  //  Add the remaining characters from the "right" string
  while (rightIndex < sortedRight.Length)
  {
    char nextChar = sortedRight[rightIndex++];
    if (lastChar == nextChar) continue;

    theResult.Append(nextChar);
    lastChar = nextChar;
  }
  theResult.Append(sortedRight, rightIndex, sortedRight.Length - rightIndex);

  return (theResult.ToString());
}

I hope that makes more sense.

Answer 3

You don't need to 2 char arrays. The System.String data type has a built-in indexer by position that returns the char from that position, so you could just loop through from 0 to (String.Length - 1). If you're more interested in speed than optimizing storage space, then you could make a HashSet for the one of the strings, then make a second HashSet which will contain your final result. Then you iterate through the second string, testing each char against the first HashSet, and if it exists then add it the second HashSet. By the end, you already have a single HashSet with all the intersections, and save yourself the pass of running through the Hashtable looking for ones with a non-zero value.

EDIT: I entered this before all the comments on the question about not wanting to use any built-in containers at all

Answer 4

here's how I would do this. It's still O(N) and it doesn't use a hash table but instead one int array of length 26. (ideally)

1. make an array of 26 integers, each element for a letter of the alphebet. init to 0's.
2. iterate over the first string, decrementing one when a letter is encountered.
3. iterate over the second string and take the absolute of whatever is at the index corresponding to any letter you encounter. (edit: thanks to scwagner in comments)
4. return all letters corresponding to all indexes holding value greater than 0.

still O(N) and extra space of only 26 ints.

of course if you're not limited to only lower or uppercase characters your array size may need to change.

Answer 5

"with each letter represented at most once"

I'm assuming that this means you just need to know the intersections, and not how many times they occurred. If that's so then you can trim down your algorithm by making use of yield . Instead of storing the count and continuing to iterate the second string looking for additional matches, you can yield the intersection right there and continue to the next possible match from the first string.