题解 | #两个链表的第一个公共结点#

python3，除了双指针之外的解法。
先计算长度，再决定谁先走多少步。空间复杂度就是O(max(m,n))

class Solution:
    def FindFirstCommonNode(self, pHead1, pHead2):
        a, b = 0, 0
        def length(head):
            if not head:
                return 0
            return 1 + length(head.next)
        a = length(pHead1)
        b = length(pHead2)
        a = a - b
        if a > 0:
            while pHead1 != pHead2:
                if a <= 0:
                    pHead2 = pHead2.next
                a = a - 1
                pHead1 = pHead1.next
        else:
            while pHead1 != pHead2:
                if a >= 0:
                    pHead1 = pHead1.next
                a = a + 1
                pHead2 = pHead2.next
        return pHead1