題目：

輸入兩個連結串列，找出它們的第一個公共結點。

解答：

第一種，蠻力法，時間複雜度為O(mn)，不需要輔助空間

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def findFirstCommonNode(self, headA, headB):
        """
        :type headA, headB: ListNode
        :rtype: ListNode
        """
 

        common = None
        a = headA
        while(a):
            b = headB
            while(b):
                if b == a:
                    return a
                b = b.next
            a = a.next
        return common    

第二種，使用棧，時間和空間複雜度為O(m+n)

# Definition for singly-linked list.
 

# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution(object):
    def findFirstCommonNode(self, headA, headB):
        """
        :type headA, headB: ListNode
        :rtype: ListNode
        """
        stackA, stackB = [], []
        a,
 
 b = headA, headB
        while(a):
            stackA.append(a)
            a = a.next
        while(b):
            stackB.append(b)
            b = b.next
        last = None
        while(stackA and stackB):
            ag, bg = stackA.pop(), stackB.pop()
            if ag == bg:
                last = ag
        return last

第三種，時間複雜度為O(m + n)，不需要輔助空間

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution(object):
    def findFirstCommonNode(self, headA, headB):
        """
        :type headA, headB: ListNode
        :rtype: ListNode
        """
        if not headA or not headB:
            return None
        a, b = headA, headB
        lenA, lenB = 0, 0
        while(a):
            lenA += 1
            a = a.next
        while(b):
            lenB += 1
            b = b.next
        last = None
        cha = lenA - lenB
        a, b = headA, headB
        if(cha > 0):
            while(cha > 0):
                a = a.next
                cha -= 1
        if(cha < 0):
            cha = -cha
            while(cha > 0):
                b = b.next
                cha -= 1
            
        while(a and b):
            if a == b:
                return a
            else:
                a, b = a.next, b.next