题目

输入两个链表，找出它们的第一个公共结点。

思路

• 1.先找到两个链表的长度length1、length2

• 2.让长一点的链表先走length2-length1步，让长链表和短链表起点相同

• 3.两个链表一起前进，比较获得第一个相等的节点

• 时间复杂度O(length1+length2) 空间复杂度O(0)

代码

function FindFirstCommonNode(pHead1, pHead2) {
      if (!pHead1 || !pHead2) { return null; }
      // 获取链表长度
      let length1 = getLength(pHead1);
      let length2 = getLength(pHead2);
      // 长链表先行
      let lang, short, interval;
      if (length1 > length2) {
        lang = pHead1;
        short = pHead2;
        interval = length1 - length2;
      } else {
        lang = pHead2;
        short = pHead1;
        interval = length2 - length1;
      }
      while (interval--) {
        lang = lang.next;
      }
      // 找相同节点
      while (lang) {
        if (lang === short) {
          return lang;
        }
        lang = lang.next;
        short = short.next;
      }
      return null;
    }

    function getLength(head) {
      let current = head;
      let result = 0;
      while (current) {
        result++;
        current = current.next;
      }
      return result;
    }