
Sorting Algorithm

This is a sorting algorithm.   It may be applied to a set of data in order to sort it.    
For comparing various sorts, see compare sorts.  
For other sorting algorithms,   see sorting algorithms,   or:

O(n logn) sorts
Heap sort |
Merge sort |
Patience sort |
Quick sort
O(n log2n) sorts
Shell Sort
O(n2) sorts
Bubble sort |
Cocktail sort |
Cocktail sort with shifting bounds |
Comb sort |
Cycle sort |
Gnome sort |
Insertion sort |
Selection sort |
Strand sort
other sorts
Bead sort |
Bogo sort |
Common sorted list |
Composite structures sort |
Custom comparator sort |
Counting sort |
Disjoint sublist sort |
External sort |
Jort sort |
Lexicographical sort |
Natural sorting |
Order by pair comparisons |
Order disjoint list items |
Order two numerical lists |
Object identifier (OID) sort |
Pancake sort |
Quickselect |
Permutation sort |
Radix sort |
Ranking methods |
Remove duplicate elements |
Sleep sort |
Stooge sort |
[Sort letters of a string] |
Three variable sort |
Topological sort |
Tree sort


Task

Given an integer   n,   return   1──►n   (inclusive)   in lexicographical order.



Show all output here on this page.



Example

Given   13,

return:   [1,10,11,12,13,2,3,4,5,6,7,8,9].

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
int compareStrings(const void *a, const void *b) {
    const char **aa = (const char **)a;
    const char **bb = (const char **)b;
    return strcmp(*aa, *bb);
}
 
void lexOrder(int n, int *ints) {
    char **strs;
    int i, first = 1, last = n, k = n, len;
    if (n < 1) {
        first = n; last = 1; k = 2 - n;
    } 
    strs = malloc(k * sizeof(char *));
    for (i = first; i <= last; ++i) {
        if (i >= 1) len = (int)log10(i) + 2;
        else if (i == 0) len = 2;
        else len = (int)log10(-i) + 3; 
        strs[i-first] = malloc(len);
        sprintf(strs[i-first], "%d", i);
    }
    qsort(strs, k, sizeof(char *), compareStrings);
    for (i = 0; i < k; ++i) {
        ints[i] = atoi(strs[i]);
        free(strs[i]);
    }
    free(strs);
}
 
int main() {
    int i, j, k, n,  *ints;
    int numbers[5] = {0, 5, 13, 21, -22};
    printf("In lexicographical order:\n\n");
    for (i = 0; i < 5; ++i) {
        k = n = numbers[i];
        if (k < 1) k = 2 - k;
        ints = malloc(k * sizeof(int));
        lexOrder(n, ints);
        printf("%3d: [", n);
        for (j = 0; j < k; ++j) {
            printf("%d ", ints[j]);
        }
        printf("\b]\n");
        free(ints);
    }
    return 0;
}