LeetCode | 160. 相交链表

O链接

• 我们这里有两个问题，一是判断是否相交，二是找交点

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思路一： 暴力求解
A链表所有节点依次取B链表找一遍（时间复杂度是O(N^2)）

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    struct ListNode *headBbak = headB;
    while(headA != NULL)
    {
        while(headB != NULL)
        {
            if(headA == headB)
                return headA;
            headB = headB->next;
        }
        headA = headA->next;
        headB = headBbak;
    }
    return NULL;
}

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思路二： 优化->要求时间复杂度到O(N)

分别找到尾结点，尾节点地址相同就相交
分别找到尾结点，尾节点地址不相同就不相交

分别求出A和B的长度~~
长的先走差距步，再同时走，第一个相同的就是交点

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struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    struct ListNode* curA = headA, *curB = headB;
    //计算长度
    int lenA = 1,lenB = 1;
    while(curA->next)
    {
        lenA++;
        curA = curA->next;
    }
    while(curB->next)
    {
        lenB++;
        curB = curB->next;
    }
    //判断相交
    while(curA != curB)
    {
        return NULL;
    }
    int n = abs(lenA - lenB);
    //假设headA长，headB短
    struct ListNode* longList = headA, *shortList = headB;
    if(lenA < lenB)
    {
        longList = headB;
        shortList = headA;
    }
    //让长链表先走n步
    while(n--)
    {
        longList = longList->next;
    }
    //两个链表同时走，直到遇到相同的节点
    while(longList != shortList)
    {
        longList = longList->next;
        shortList = shortList->next;
    }
    //相遇了，直接return
    return longList;
}