Two methods for the full array of strings

Enter a string that contains different characters. The program outputs all permutations (all permutations) of the string.

voidFChar*STR,intLenintN) {       inti; Chartmp; Char*p = (Char*)malloc(len+1); if(n==len-1) {printf ("%s\n", str); }Else{            for(i=n;i<len;i++) {strcpy (P,STR); TMP= * (str+N); * (str+n) = * (str+i); * (str+i) =tmp; F (Str,len,n+1);           strcpy (STR,P); }       }        Free(P); }   intMainintargcChar**argv) {       CharStr[] ="XYZ"; F (str,3,0); printf ("\ n"); return 0; }

For example: Given the string "XYZ", the program outputs:
Xyz
Xzy
Yxz
zz[
yy[

Zxy

Also enter a string, where you can include a repeating string, and the output is fully arranged.

#include <stdio.h>#include<string.h>#include<memory.h>intM//Record string lengthintN//the number of character types in the record stringCharmap[ the];//What kinds of characters are recordedintcount[ the];//record how many of each character voidMake_map (Char*STR)//information about the statistics string{     ints[ the]; inti; memset (s),0,sizeof(s)); memset (Count,0,sizeof(count)); M=strlen (str);  while(*str) {s[*str]++; STR++; } N=0;  for(i=0;i< the; i++)         if(S[i]) {Map[n]=i; Count[n]=S[i]; N++; } }  intstack[ +];//recursively uses stacks, and records the current generated permutations voidFind (intDepth//recursive backtracking for full-array generation{         intk=0; if(depth==m) {inti;  for(i=0; i<depth;i++) {Putchar (map[stack[i]); //printf ("%d\n", k);} putchar ('\ n'); }         Else         {             inti;  for(i=0; i<n;i++)                 if(Count[i]) {stack[depth]=i; Count[i]--; Find (Depth+1); Count[i]++; }         } }  intMain () {Charstr[ +];     Gets (str);     Make_map (str); Find (0); return 0;}

Note: If you encounter a,s[*str]++ representing s[65]++, you encounter B, then s[' B ']++
Finally, s[' A '],s[' B '],s[' C ' ... The value is a, B 、...... The number of occurrences of these letters

Two methods for the full array of strings