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Partition LinkedList.c
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Partition LinkedList.c
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Problem:
https://www.interviewbit.com/courses/programming/topics/linked-lists/problems/partitionlist/?ref=dash-reco
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.
Solution:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*
* typedef struct ListNode listnode;
*
* listnode* listnode_new(int val) {
* listnode* node = (listnode *) malloc(sizeof(listnode));
* node->val = val;
* node->next = NULL;
* return node;
* }
*/
/**
* @input A : Head pointer of linked list
* @input B : Integer
*
* @Output head pointer of list.
*/
int listCount(listnode* A){
int a=0;
while(A!=NULL){
a++;
A = A->next;
}
return a;
}
listnode* partition(listnode* A, int B) {
int d = listCount(A);
if(d==1){
return A;
}
int i=0;
listnode *end = A, *temp= A;
while(end->next != NULL){
end = end->next;
}
listnode *prev = A;
while(i<d && A!=NULL){
if(temp->val >= B){
/*if(temp==end){
break;
}*/
if(prev == temp && temp->val >= B){
listnode *create = listnode_new(temp->val);
end->next = create;
end = create;
prev = temp->next;
free(temp);
A = temp = prev;
}else{
listnode *create = listnode_new(temp->val);
end->next = create;
end = create;
prev->next = temp->next;
free(temp);
temp = prev->next;
}
}else{
if(temp!=prev){
prev = prev->next;
}
temp = temp->next;
}
i++;
}
return A;
}