careercup-Recursive and dynamic programming 9.5

9.5 Write a method that determines all permutation combinations of a string.

Similar leetcode:permutations

Solution:

As with many recursive problems, the simple construction method works very well. Suppose there is a string s, expressed as a sequence of characters a1a2a...an.

Termination Condition: N=1

S=A1, there is only one permutation combination, that is, the string A1

Situation: n=2

S=A1A2 There are two permutations of A1A2 and A2A1

Situation: N

For the general case, we just need to repeat this step. That is, the solution of F (n-1) has been obtained, and then as long as the an is inserted into any position of these strings.

C + + code implementation:

#include <iostream>#include<vector>#include<string>using namespaceStd;vector<string> getperms (stringstr) {    if(Str.empty ())returnvector<string>(); intn=str.length (); Vector<string>Res; Res.push_back (string("")); Vector<string>R; inti,j;  for(i=0; i<n;i++)    {        CharC=Str[i]; R=Res;        Res.clear ();  for(j=0; J<r.size (); j + +)        {           stringtmp=R[j]; intindex=0;  while(Tmp.begin () +index!=Tmp.end ()) {Tmp.insert (Tmp.begin ()+index,c);               Res.push_back (TMP); TMP=R[j]; Index++;           } tmp.push_back (c);        Res.push_back (TMP); }    }    returnRes;}intMain () {stringStr="ABC"; Vector<string> res=getperms (str);  for(auto s:res) cout<<s<<Endl;}

careercup-Recursive and dynamic programming 9.5