逆转单链表

1. 非递归实现

(1) (没有头节点的)其中 List 跟 Position 是结构指针;

typedef struct Node {
    int value;
    struct Node *next;
} *List;

typedef List Position;

List
ReverseList( List L ) {  // Assuming no header.
    if( L == NULL )
        return L;
    Position PreviousPos, CurrentPos, NextPos;

    PreviousPos = NULL;
    CurrentPos = L;
    NextPos = L->next;
    while( NextPos != NULL ) {
        CurrentPos->next = PreviousPos;
        PreviousPos = CurrentPos;
        CurrentPos = NextPos;
        NextPos = NextPos->next;
    }
    CurrentPos->next = PreviousPos;

    return CurrentPos;
}

(2) (有头结点的)

typedef struct Node {
    int value;
    struct Node *next;
} *List;

typedef List Position;

List
ReverseList( List L ) {  // Assumed header
    if( L == NULL )
        return L;
    if( L->next == NULL )
        return L;
    Position PreviousPos, CurrentPos, NextPos;

    PreviousPos = NULL;
    CurrentPos = L->next;
    NextPos = CurrentPos->next;
    while( NextPos != NULL ) {
        CurrentPos->next = PreviousPos;
        PreviousPos = CurrentPos;
        CurrentPos = NextPos;
        NextPos = NextPos->next;
    }
    CurrentPos->next = NextPos;
    L->next = CurrentPos;

    return L;
}

2. 递归实现

p1 和 p2 指针分别指向当前递归子链表 list1 的第一个和第二个结点。然后对以 p2 为首结点的子链表 list2 进行递归逆转;则 p2 节点将成为 list2r 逆转后的尾结点,而此时函数返回的头结点将是原 list2 的尾结点(如下图)。最后我们只要把 p2 的 next 指向 p1 就可以了。

image


image

(1) 递归实现

typedef struct Node {
    int value;
    struct Node *next;
} *List;

typedef List Position;

List
ReverseList( List L ) {   // Assuming no header
    if( L == NULL )
        return L;
    if( L->next == NULL )
        return L;

    Position p1, p2;

    p1 = L;
    p2 = p1->next;
    L = ReverseList( p2 );
    p2->next = p1;
    p1->next = NULL;

    return L;
}
posted @ 2016-10-30 09:00  夏夜星空晚风  阅读(364)  评论(0编辑  收藏  举报