26.LeetCode之相交链表

编写一个程序，找到两个单链表相交的起始节点。

题目链接：https://leetcode-cn.com/problems/intersection-of-two-linked-lists/

相交链表解题思路详解：https://blog.youkuaiyun.com/dawn_after_dark/article/details/73864643

C代码：双指针(每个指针都遍历了AB两个链表)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
/*
*定义两个指针, 第一轮让两个到达末尾的节点指向另一个链表的头部, 最后如果相遇则为交点(在第一轮移动中恰好抹除了长度差)
*两个指针等于移动了相同的距离, 有交点就返回, 无交点就是各走了两条指针的长度
*/
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    if( headA==NULL || headB==NULL )
        return NULL;
    struct ListNode *pA = headA, *pB = headB;
    // 在这里第一轮体现在pA和pB第一次到达尾部会移向另一链表的表头, 
    //而第二轮体现在如果pA或pB相交就返回交点, 不相交最后就是null==null
    while( pA != pB )
    {
        pA = pA==NULL ? headB : pA->next;
        pB = pB==NULL ? headA : pB->next;
    }
    return pA;
}

C代码：双指针 [B链表变环，验证A链表是否有环。若有环，由于快指针追上了慢指针，此时将慢指针指向A链表的头节点，使得slow和fast指针与交点距离相同，齐头并进(slow==fast)即可找出交点]

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
//看不懂就结合leedcode的图走一遍
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    if( headA==NULL || headB==NULL )
        return NULL;
    struct ListNode *last = headB;
    while( last->next != NULL )
        last = last->next;
    last->next = headB;
    struct ListNode *slow=headA, *fast=headA; //注意此时的slow和fast必须从同一起点开始
    while( fast!=NULL && fast->next!=NULL ) //链表还没遍历完
    {
        slow = slow->next;
        fast = fast->next->next;
        if( slow == fast ) //fast指针追上slow指针，故有环
        {
            slow = headA; //此时消除了slow,fast之间距离差距，使得slow,fast与交点距离相同
            while( slow != fast ) //slow==fast时，指向相交点
            {
                slow = slow->next;
                fast = fast->next;
            }
            last->next = NULL;
            return fast;
        }
    } //退出while循环，说明无环，则两链表不相交
    last->next = NULL;
    return NULL;
}

C代码：双指针(长链表的长度减去短链表的长度，得到相差的长度，之后让长链表从头结点开始走过这个长度，与短链表同时向后走，若指针相等，则链表相交；若走到NULL，则链表不相交)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    struct ListNode *pA = headA, *pB = headB;;
    struct ListNode *tmp = NULL;
    int lenght_A = 0 ,lenght_B = 0;

    while( pA != NULL ) //计算链表A,B的长度
    {
        pA = pA->next;
        lenght_A++;
    }
    while( pB != NULL ) 
    {
        pB = pB->next;
        lenght_B++;
    }
    
  //使pA,pB分别指向距离A,B链表的尾结点相同的结点
    pA = headA;
    pB = headB;
    if (lenght_A > lenght_B) {
        int gap = lenght_A - lenght_B;
        while( gap-- ) 
            pA = pA->next;
    } else {
        int gap = lenght_B - lenght_A;
        while( gap-- ) 
            pB = pB->next;
    }
    
    //遍历找出相交结点
    while ( pA && pB ) 
    {
        if( pA ==pB )
        {
            return pA;
        }
        pA = pA->next;
        pB = pB->next;
    }
    return NULL;
}

注：若题目改为验证两个链表是否相交，则可以变换思路为使B链表的尾结点指针指向该链表的头节点，然后使用快慢指针的方法验证A链表是否有环。若有，则A,B链表相交，否则不相交

缺点：修改了原链表结构

C代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
bool getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    if( headA==NULL || headB==NULL )
        return false;
    struct ListNode *last = headB;
    while( last->next != NULL )
        last = last->next;
    last->next = headB;
    struct ListNode *slow=headA, *fast=headA->next;
    while( fast!=NULL && fast->next!=NULL ) //链表还没遍历完
    {
        slow = slow->next;
        fast = fast->next->next;
        if( slow == fast ) //fast指针追上slow指针，故有环
            return true;
    }
    return false;
}