234 Palindrome Linked List

234 Palindrome Linked List

这道题如果不要求space O(1) 代码写起来就相当简单。 但是要求 O(1),唯一的方法就是reverse 前半部分linked List 在做比较， 稍微注意长度为奇偶时一点小小的区别。写起来还是很费劲的

class Solution:
    # @param {ListNode} head
    # @return {boolean}
    def isPalindrome(self, head):
        dummy = ListNode(0)
        dummy.next = head
        p1, p2, length = dummy, dummy,0
        while p2.next != None:
            p1 = p1.next
            p2 = p2.next
            length += 1
            if p2.next == None:
                break
            p2 = p2.next
            length += 1
        p2 = p1.next
        p1 = self.reverse(head, p1)
        if length % 2 == 1:
            p1 = p1.next
        while p2 != None:
            if p1.val != p2.val:
                return False
            p1, p2 = p1.next, p2.next
        return True

    def reverse(self, head, p1):
        if head == p1:
            return head
        if head == None:
            return
        net = head.next
        head.next = None
        rest = self.reverse(net,p1)
        net.next = head
        return rest