160.链表交点

题目

Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:

分析

• 由于连个链表的长度不同，没法直接比较，需要移动指针使两个指针分别在两个链表的相同索引处
• 可以使用长度相减法使指针处于相同索引处，也可以使用补齐法使指针处于相同索引处

代码如下

补齐法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    struct ListNode *p, *q;
    p = headA;
    q = headB;
    //补齐思想，较短的链表先到达末尾，然后跳到较长链表的头节点,这样就相当于对短链表进行了补齐
    while(p != q) { 
        p = (p ? p->next : headB);
        q = (q ? q->next : headA);
    }
    return p;
}

相减法

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    
    struct ListNode *current_node_a = headA;
    struct ListNode *current_node_b = headB;
    int length_a = 1;
    int length_b = 1;
    if (headA == NULL || headB == NULL) return NULL;
    while (current_node_a->next != NULL) {
        current_node_a = current_node_a->next;
        length_a++;
    }

    while (current_node_b->next != NULL) {
        current_node_b = current_node_b->next;
        length_b++;
    }
    current_node_a = headA;
    current_node_b = headB;
    
    if (length_a < length_b) {
        int temp = length_b - length_a;
        while(temp--) current_node_b = current_node_b->next;
    }
    if (length_a > length_b) {
        int temp = length_a - length_b;
        while(temp--) current_node_a = current_node_a->next;
    }
    if (current_node_a == current_node_b) return current_node_a;
    while(current_node_a->next != NULL) {
        if (current_node_a->next == current_node_b->next) return current_node_a->next;
        current_node_a = current_node_a->next;
        current_node_b = current_node_b->next;
    }
    return NULL;

}