[LeetCode]83. Intersection of Two Lists两条单链表的交点

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

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解法：两条单链表如果在某个节点相交，那么从这个节点开始，往后的所有节点都是一样的（因为从这个相交节点开始，所有节点的next指针都是一样的）。因此相交节点往后的长度是一样的。因此如果两条链表长度不一样，需要先移动长链表的指针，使得从长度一样的位置处开始一一比较。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *currA = headA, *currB = headB;
        int diffLen = 0; // 两条链表的长度差
        while (currA != NULL && currB != NULL) {
            currA = currA->next;
            currB = currB->next;
        }
        if (currA != NULL || currB != NULL) {
            while (currA != NULL) {
                ++diffLen;
                currA = currA->next;
            }
            while (currB != NULL) {
                --diffLen;
                currB = currB->next;
            }
        }
        currA = headA;
        currB = headB;
        if (diffLen != 0) { // 存在长度差则先移动到相同长度节点处
            while (diffLen-- > 0) currA = currA->next;
            while (++diffLen < 0) currB = currB->next;
        }
        while (currA != NULL && currB != NULL && currA->val != currB->val) {
            currA = currA->next;
            currB = currB->next;
        }
        return currA;
    }
};

或者先分别算出两条链表长度再比较：

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *currA = headA, *currB = headB;
        int sizeA = 0, sizeB = 0;
        while(currA != NULL) {
            ++sizeA;
            currA = currA->next;
        }
        while(currB != NULL) {
            ++sizeB;
            currB = currB->next;
        }
        currA = headA;
        currB = headB;
        if(sizeA != sizeB) {
            int diff = sizeA - sizeB;
            while(diff-- > 0) currA = currA->next;
            diff = sizeB - sizeA;
            while(diff-- > 0) currB = currB->next;
        }
        while(currA != NULL && currB != NULL && currA->val != currB->val) {
            currA = currA->next;
            currB = currB->next;
        }
        return currA;
    }
};