一道面试题的实现

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#面试 #null #struct #算法

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本文介绍了一种单链表结构中将偶数位置元素进行倒排的算法实现，采用4n时间复杂度和O(1)空间复杂度，通过两个阶段的反转操作完成目标，并附带了C语言代码示例。

问题：单链表结构，把偶数位置的链表倒排，例如

1->2->3->4->5->6->7->null

转化为

1->6->3->4->5->2->7->null

当场设计了一个算法，时间4n，空间O(1)。感觉还不错，被面试官说了代码不规范。回来后重写了一遍，发现确实有很多漏洞。

代码附下，请有经验者指点！

#include <stdio.h> #define MAXSIZE 100 typedef struct Head { int content; struct Head* next; } Head; void change(Head* head) { int length=0, midPos; Head *pt1, *pt2, *temp, *mp, *mlp, *mrp; //point to midPosition, midPosition-1, midPosition+1 //calculate the length temp = head; while (temp) { ++length; temp = temp->next; } if ( length < 4 ) return; midPos = length/2+1; //reverse pointer to before midPos and record the point. pt1 = head; pt2 = pt1->next; pt1->next = NULL; while (--midPos > 1) { temp = pt2->next; pt2->next = pt1; pt1 = pt2; pt2 = temp; } mp = pt2; mlp = pt1; mrp = pt2->next; pt1 = mlp; pt2 = mrp; while ( pt1 != head && pt2 != head ) { temp = pt1->next; pt1->next = pt2->next; pt2->next = temp; pt1 = pt1->next; pt2 = pt2->next; } if ( pt1 == head && pt2 == NULL ) { } else if ( pt1 == head && pt2 != NULL ) { pt2->next = NULL; } else if ( pt2 == head && pt1 == NULL ) { temp = mlp; mlp = mrp; mrp = temp; } else if ( pt2 == head && pt1 != NULL ) { temp = mlp; mlp = mrp; mrp = temp; pt1->next = NULL; } mp->next = mrp; pt1 = mlp; pt2 = mlp->next; mlp->next = mp; while (pt2) { temp = pt2->next; pt2->next = pt1; pt1 = pt2; pt2 = temp; } } int main() { Head* head = NULL, *temp; int i; for (i=0; i<MAXSIZE; ++i) { temp = (Head*) malloc(sizeof(Head)); temp->content = MAXSIZE-i; temp->next = head; head = temp; } change(head); temp = head; while (temp) { printf("%i/n", temp->content); temp = temp->next; } getchar(); return 0; }