题解 | #链表中的节点每k个一组翻转#

// 复杂问题分两步解决
/**
 * 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 <stdio.h>
struct ListNode* reverseRange(struct ListNode* head,int m,int n){
    struct ListNode* p;
    struct ListNode* pre;
    int i;
    p = head;
    pre = NULL;
    //到达指定的位置
    for (i = 1;i<m;i++){
        pre = p;
        p = p->next;
    }

    struct ListNode* temp;
    struct ListNode* temp_ne;
    struct ListNode* q;
    temp = p->next;
    temp_ne = temp->next;
    
    q = p;
    for(i = 0;i<n-m;i++){
            temp->next = p;
            p = temp;
            temp = temp_ne;
            temp_ne = temp->next;
        }
        q->next = temp;
    //特殊情况
    if(pre == NULL){
        return p;
    }else{
        pre->next = p;
        return head;
    }
}


struct ListNode* reverseKGroup(struct ListNode* head, int k ) {
    struct ListNode* p_head;
    struct ListNode* resul_l;
    int count = 1;
    int length = 1;
    p_head = head;
    resul_l = head;
    while(p_head!=NULL){
        p_head = p_head->next;
        length++;
        printf("%d",length);
    }
    while(count<length){
        if(count % k == 0){
            resul_l = reverseRange(resul_l,count-k+1,count);
        }
        count ++ ;
    }
    return resul_l;
}