一道面試題的實現

問題：單鏈表結構，把偶數位置的鏈表倒排，例如

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

轉化爲

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

 

當場設計了一個算法，時間4n，空間O(1)。感覺還不錯，被面試官說了代碼不規範。回來後重寫了一遍，發現確實有很多漏洞。

 

代碼附下，請有經驗者指點！

 

#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; }