# 0ms C solution, simple and easy to comprehend.

• Since the List given by OJ doesn't have a leading node, I create a sentinel node for convenient.

``````struct ListNode* reverseBetween(struct ListNode* head, int m, int n) {
if (head == NULL || m == n)

struct ListNode sentinel;  /* Create a sentinel node for the list */
struct ListNode *m_preceding = &sentinel;  /* Position of the node precedes m */
int count = 0;  /* Counter */
int distance = n - m;  /* Times needed to reverse the internal list [m,n] */

while (count++ < m-1)  /* Find the node precedes m */
m_preceding = m_preceding->next;

struct ListNode *p = m_preceding->next;  /* p points to node m */
struct ListNode *q = p->next;  /* q points to the node next to m */

for (count = 0; count < distance; count++)  /* Step1, reverse the list between [m,n] iteratively */
{
struct ListNode *temp = q->next;  /* Save the address of the node succeeds n */
q->next = p;  /* Successor points to the predecessor */
p = q;  /* Move p downward one spot */
q = temp;  /* Move q doward one sopt */
}
m_preceding->next->next = q;  /* Step2, fix node m pointing to the node after n originally */
m_preceding->next = p;  /* and the node precedes m pointing to n */

return sentinel.next;
}
``````

If you have any better advises to improve this program, please let me know.
Thanks, xiyu.

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