链表

1 合并两个链表

class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        # write code here
        p1 = pHead1
        p2 = pHead2
        p3 = ListNode(0)
        p = p3
        while p1 and p2:
            if p1.val >= p2.val:
                p3.next = p2
                p2 = p2.next
            else:
                p3.next = p1
                p1 = p1.next
            p3 = p3.next
            
        if p1:
            p3.next = p1
        else:
            p3.next = p2
        return p.next

2 链表判环 并返回入环节点的值

class ChkLoop:
    def chkLoop(self, head, adjust):
        # write code here
        slow = head
        fast = head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
            if fast == slow:
                fast = head
                while fast != slow:
                    fast = fast.next
                    slow = slow.next
                return fast.val
        return -1

3 两个无环单链表是否相交

class CheckIntersect:
    def chkIntersect(self, headA, headB):
        # write code here
        lenA = 0
        lenB = 0
        pA = headA
        pB = headB
        while pA:
            pA = pA.next
            lenA += 1
        while pB:
            pB = pB.next
            lenB += 1
        pA = headA
        pB = headB
        while lenA != lenB:
            if lenA > lenB:
                pA = pA.next
                lenA -= 1
            else:
                pB = pB.next
                lenB -= 1
        while pA:
            if pA == pB:
                return True
            pA = pA.next
            pB = pB.next
        return False

4 合并两个有序链表

class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        # write code here
        p1 = pHead1
        p2 = pHead2
        p3 = ListNode(0)
        p = p3
        while p1 and p2:
            if p1.val >= p2.val:
                p3.next = p2
                p2 = p2.next
            else:
                p3.next = p1
                p1 = p1.next
            p3 = p3.next
            
        if p1:
            p3.next = p1
        else:
            p3.next = p2
        return p.next

5 链表排序

class Solution(object):
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        return self.mergeSort(head)
    
    def mergeSort(self, head):
        if head is None or head.next is None:
            return head
        
        fast = slow = head
        while fast and fast.next:
            fast = fast.next.next
            breakN = slow
            slow = slow.next
        breakN.next = None
        l1 = self.mergeSort(head)
        l2 = self.mergeSort(slow)
        return self.merge(l1, l2)
    
    def merge(self, l1, l2):
        if l1 is None:
            return l2
        if l2 is None:
            return l1
        if l1.val <= l2.val:
            l1.next = self.merge(l1.next, l2)
            return l1
        else:
            l2.next = self.merge(l2.next, l1)
            return l2