LintCode: 两个链表的交叉

LintCode: 两个链表的交叉

最容易想到的当然是类似冒泡算法，代码：

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

class Solution:
    # @param headA: the first list
    # @param headB: the second list
    # @return: a ListNode
    def getIntersectionNode(self, headA, headB):
        # Write your code here
        if headA == None or headB == None:
            return None
        p1 = headA
        p2 = headB
        while p1.next != None:
            while p2.next != None:
                if p1 == p2:
                    return p1
                p2 = p2.next
            p2 = headB
            p1 = p1.next
        return None

不过不出意外地。。。超时了。

另一种方案，先找到较长的链表，然后将其向后移动|len(A) - len(B)|的距离，这样两个链表就具有相同的长度了，就不必再进行冒泡比较了，复杂度也就降为n。

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

class Solution:
    # @param headA: the first list
    # @param headB: the second list
    # @return: a ListNode
    def getIntersectionNode(self, headA, headB):
        # Write your code here
        if headA == None or headB == None:
            return None
        p1 = headA
        p2 = headB
        len_a = 0
        len_b = 0
        while p1.next != None:
            len_a += 1
            p1 = p1.next
        while p2.next != None:
            len_b += 1
            p2 = p2.next
        len_c = abs(len_a - len_b)
        p1 = headA
        p2 = headB
        if len_a >= len_b:
            for i in range(len_c):
                p1 = p1.next
        else:
            for i in range(len_c):
                p2 = p2.next
        while p1 != None and p2 != None:
            if p1 == p2:
                return p1
            p1 = p1.next
            p2 = p2.next
        return None