Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

#include<stdio.h>#include<stdlib.h>typedef struct ListNode {   int val;   struct ListNode *next;}ListNode;ListNode *partition(ListNode *head, int x) {    ListNode *less = NULL, *bigger = NULL, *p;    ListNode *less_index , *bigger_index;    int less_cnt = 0, bigger_cnt = 0;    if(head == NULL) return NULL;    for(p = head; p != NULL; p = p->next) {        if(p->val < x) {            if(less_cnt == 0) less_index = less = p;             else {                less_index->next = p;                less_index = p;            }            less_cnt++;        }        else {            if(bigger_cnt == 0) bigger_index = bigger = p;            else {                bigger_index->next = p;                bigger_index = p;            }            bigger_cnt++;        }    }    if(less != NULL) less_index->next = bigger;    else less = bigger;    if(bigger != NULL) bigger_index->next = NULL;    return less;}void main() {    ListNode *head, *p1, *p2;    int i = 0;    p1 = p2 = (ListNode *)malloc(sizeof(ListNode));    scanf("%d", &p1->val);    while(p1->val != -1) {        i++;        if(i == 1) head = p1;        else p2->next = p1;        p2 = p1;        p1 = (ListNode *)malloc(sizeof(ListNode));        scanf("%d", &p1->val);    }    p2->next = NULL;    for(p1 = head; p1 != NULL; p1 = p1->next) printf("%d ", p1->val);    printf("\n");    partition(head, 0);}