Enter a string, arrange to print out all of the characters in this string,



You can return the string array in any order, but it can't have a repeating element,



Example:

Input: s="ABC" output: [" ABC ", "acb", "America", "bca", "cab", "cba"]


Title url: https://leetcode-cn.com/problems/zi-fu-chuan-de-pai-lie-lcof/solution/


I run error code

The class Solution (object) :
LST=[]
Def permutation (self, s) :
"" "
: type s: STR
Rtype: List (STR)
"" "
Self. Perm (list (s), and 0, len (s))
Return a list (sorted (set (self. LST)))
Def perm (self, s, k, m) :
If k==m:
The self. The LST. Append (". "the join (s))
The else:
For I in range (k, m) :
S [I], [k] s=s [k], [I] s
Self. Perm (s, k + 1, m)
S [I], [k] s=s [k], [I] s

CodePudding user response:

Fyi:

 # include & lt; Stdio. H> 
# include & lt; String. H> 
# include & lt; Stdlib. H> 
int m;//record the length of the string 
int n;//record the number of types of characters in a string 
Char map [256].//record is what kind of character 
Int count [256];//record each character how many 
Int stack [1000];//recursive stack, and record the arrangement of the current generation 
Void Make_Map (char * STR) {//statistics the related information of the string 
Int s [256]; 
int i; 
Memset (s, 0, sizeof (s)); 
Memset (count, 0, sizeof (count)); 
M=strlen (STR); 
While (* STR) {
S [* STR] + +; 
Str++; 
} 
N=0; 
For (I=0; i<256; I++) 
If (s) [I] {
The map [n]=I; 
The count [n] [I]=s; 
n++; 
} 
} 
Void the Find (int the depth) {//recursion backtracking generating permutations 
If (the depth==m) {
int i; 
For (I=0; iputchar('\n'); 
} else {
int i; 
For (I=0; iIf (count [I]) {
Stack [the depth]=I; 
The count [I] -; 
Find the depth (+ 1); 
The count [I] + +; 
} 
} 
} 
Void main (int arg c, char * * argv) {
If (argc<2) {
Printf (" % s to produce the whole arrangement of string \ n ", argv [0]). 
return; 
} 
Make_Map (argv [1]); 
Find (0); 
}