题解 | #链表中的节点每k个一组翻转#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param head ListNode类
 * @param k int整型
 * @return ListNode类
 */
#include <stdlib.h>
struct ListNode* revers(struct ListNode* head, int m, int n) {

    struct ListNode* res = (struct ListNode*)malloc(sizeof(struct ListNode));
    res->next = head;
    if (head == NULL || m == n) {
        return head;
    } else {
        struct ListNode* pre = res;
        struct ListNode* cur = pre->next;
        struct ListNode* post = NULL;
        for (int i = 1; i < m; i++) {
            pre = pre->next;
        }
        cur = pre->next;
        for (int i = m; i < n; i++) {
            post = cur->next;
            cur->next = post->next;
            post->next = pre->next;
            pre->next = post;
        }
        return res->next;
    }
}

struct ListNode* reverseKGroup(struct ListNode* head, int k ) {
    // write code here
    struct ListNode*p=head;
    int i=0;
    while(p!=NULL){
        i++;
        p=p->next;
    }
    if(k>i){
        return head;
    }
    for(int n=1;n<i;n=n+k){
        if(n+k-1>i){
            break;
        }
        head=revers(head,n,n+k-1);
        
    }

    return head;
}