Algorithm: whether the characters in a short string are all contained in a long string

Algorithm title: There are two strings made up of different letters, one long and one short, long for a short for b. Design an algorithm that returns true if all characters appearing in B appear in a, otherwise false.

The following string:

String A:ABDDFDIOEGDDDFFSFAGJ

String B:DOFSJADG

Each character in string B appears in A and returns true.

The following string:

String a:aaaabbbbbbdddddd

String B:ACC

The character in string B does not appear in a and returns false.

Answer 1: Each letter in string B is iterated through in a. This answer sucks, its time complexity is O (n*m)

Answer 2: Set up a hash table, a character traversal of string A, and set the value in the hash table corresponding to each character to 1. The characters in B are then traversed, and if all characters correspond to a hash value of 1, true is returned, otherwise false is returned.

The time complexity of the answer is O (m+n), which should be the answer most interviewers want, and most people can think of.

intStrinclude (Const Char*longone,Const Char*Shortone) {      intI, allchar[ the] = {0 }; intLonglen =strlen (Longone); intShortlen =strlen (Shortone);  for(i =0; i < Longlen; i++) Allchar[longone[i]]=1;  for(i =0; i < Shortlen; i++)           if(Allchar[shortone[i]]! =1)               Break; returni = = Shortlen?1:0; }  

Algorithm: whether the characters in a short string are all contained in a long string