LeetCode092 Reverse Linked List II

详细见：leetcode.com/problems/reverse-linked-list-ii

Java Solution: github

package leetcode;/* * Reverse a linked list from position m to n. Do it in-place and in one-pass.For example:Given 1->2->3->4->5->NULL, m = 2 and n = 4,return 1->4->3->2->5->NULL.Note:Given m, n satisfy the following condition:1 ≤ m ≤ n ≤ length of list. */import tools.ListNode辅助.ListNode;;public class P092_ReverseLinkedListII {public static void main(String[] args) {ListNode head = tools.ListNode辅助.A_一维生成器(new int[] {});ListNode ans = new Solution().reverseBetween(head, 1, 1);tools.ListNode辅助.B_打印链表(ans);}/* * 题目非常容易，一次AC，好爽 * 1 ms */static class Solution {    public ListNode reverseBetween(ListNode head, int m, int n) {    if (m >= n) {    return head;    }    ListNode tra = head, one_last = null, two_first = null,    two_last = null, three_first = null, save_tra_next = null;    int count = 1;    while (tra != null) {    save_tra_next = tra.next;    if (count == m - 1) {    one_last = tra;    } else if (count == m) {    two_first = tra;    two_last = tra;    } else if (count == n + 1) {    three_first = tra;    } else if (count > m && count <= n) {    tra.next = two_first;    two_first = tra;    }    tra = save_tra_next;    count ++;    }    if (one_last != null) {    one_last.next = two_first;    }    if (two_last != null) {    two_last.next = three_first;    }    if (one_last == null) {    return two_first;    }        return head;    }}}

C Solution: github

/*    url: leetcode.com/problems/reverse-linked-list-ii    AC 0ms 73.39%*/#include <stdio.h>#include <stdlib.h>typedef struct ListNode sln;typedef struct ListNode * pln;struct ListNode {    int val;    struct ListNode * next;};pln convert_int_to_ListNode(int * arr, int n) {    pln head = NULL;    pln travel = NULL;   pln temp = NULL;    int i = 0;      if (n == 0 || n < 0) return NULL;    travel = (pln) malloc(sizeof(sln));    travel->val = *(arr + 0);    travel->next = NULL;    head = travel;    for (i = 1; i < n; i ++) {        temp = (pln) malloc(sizeof(sln));        temp->val = *(arr + i);        temp->next = NULL;        travel->next = temp;        travel = travel->next;    }    return head;}void free_ListNode(struct ListNode * l) {    struct ListNode * l1 = l, * l2 = l;    if (l == NULL) return;    while (l1 != NULL) {        l2 = l2->next;        free(l1);        l1 = l2;    }}void print_ListNode(pln l) {    while (l != NULL) {        printf("%d ", l->val);        l = l->next;    }    printf("\r\n");}pln reverseBetween(pln h, int m, int n) {    int i = 1;    pln t = h, s = NULL, p = NULL;    pln h1 = NULL, h2 = NULL;    pln e1 = NULL, e2 = NULL;    while (t != NULL) {        s = t->next;        if (i == m-1) e1 = t;        if (i == n+1) {            e2 = t;            break;        }        if (i == m) h1 = t;        if (i == n) h2 = t;        if (i > m && i <= n) {            t->next = p;        }        p = t;        t = s;        i ++;    }    if (e1 != NULL) e1->next = h2;    if (h1 != NULL) h1->next = e2;    return m == 1 ? h2 : h;}int main() {    /*        1, 2, 3, 4, 5, 6, 7, 8, 9        1, 2        1, 1        2, 4    */    int a[] = {1};    pln h = convert_int_to_ListNode(a, 1);    pln ans = reverseBetween(h, 1, 1);    print_ListNode(ans);    return 0;    }

Python Solution: github

#coding=utf-8'''    url: leetcode.com/problems/reverse-linked-list-ii    @author:     zxwtry    @email:      [e-mail]    @date:       2017年4月23日    @details:    Solution: 46ms 49.65%'''class ListNode(object):    def __init__(self, x):        self.val = x        self.next = None    def __str__(self, *args, **kwargs):        return str(self.val)def con(l):    n = [ListNode(v) for v in l]    for i in range(len(l)-1):        n[i].next = n[i+1]    return n[0]def pri(d):    print("=====================")    while d != None:        print(d.val)        d = d.next    print("=====================")class Solution(object):    def reverseBetween(self, h, m, n):        """        :type h: ListNode        :type m: int        :type n: int        :rtype: ListNode        """        t, d, p = 1, h, None        h1, h2, e1, e2 = None, None, None, None        while d != None:            if t == m-1: e1=d            if t == m: h1=d            if t == n: h2=d            if t == n+1:                e2=d                break            s = d.next            if t >= m:                d.next=p            d, t, p = s, t+1, d        if e1 != None: e1.next = h2        if h1 != None: h1.next = e2        if m == 1: return h2        return h    if __name__ == "__main__":    h = con([1, 2, 3, 4, 5])    a = Solution().reverseBetween(h, 1, 5)    pri(a)