访问单个节点的删除、链表的分化、打印两个链表的公共值

目录

访问单个节点的删除

链表的分化

打印两个链表的公共值

访问单个节点的删除

实现一个算法，删除单向链表中间的某个结点，假定你只能访问该结点。

给定带删除的头节点和要删除的数字，请执行删除操作，返回删除后的头结点。链表中没有重复数字

我的提交

（是我理解错题意了？？？）

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Remove:
def removeNode(self, pHead, delVal):
# write code here
if pHead.val == delVal:
pHead = pHead.next
return pHead
p1 = pHead
p2 = None
while p1.next != None:
p2 = p1
p1 = p1.next
if p1.val == delVal:
p2.next = p1.next
break
return pHead

参考答案

import java.util.*;

/*
public class ListNode {
int val;
ListNode next = null;

ListNode(int val) {
this.val = val;
}
}*/
public class Remove {
public boolean removeNode(ListNode pNode) {
if (pNode == null) {
return false;
}
ListNode next = pNode.next;
if (next == null) {
return false;
}
pNode.val = next.val;
pNode.next = next.next;
return true;
}
}

链表的分化

对于一个链表，我们需要用一个特定阈值完成对它的分化，使得小于等于这个值的结点移到前面，大于该值的结点在后面，同时保证两类结点内部的位置关系不变。

给定一个链表的头结点head，同时给定阈值val，请返回一个链表，使小于等于它的结点在前，大于等于它的在后，保证结点值不重复。

测试样例：
{1,4,2,5},3
{1,2,4,5}

我的提交

（审错题了）

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Divide:
def __init__(self):
self.pointer = [[], [], []]

def listDivide(self, head, val):
# write code here
for item in head:
if item < val:
self.pointer[0].append(item)
elif item == val:
self.pointer[1].append(item)
else:
self.pointer[2].append(item)
return self.pointer[0] + self.pointer[1] + self.pointer[2]

if __name__ == '__main__':
d = Divide()
result = d.listDivide({1,4,2,5},3)
print(result)

参考答案

import java.util.*;

/*
public class ListNode {
int val;
ListNode next = null;

ListNode(int val) {
this.val = val;
}
}*/
public class Divide {
public ListNode listDivide(ListNode head, int pivot) {
ListNode sH = null; // small head
ListNode sT = null; // small tail
ListNode bH = null; // big head
ListNode bT = null; // big tail
ListNode next = null; // save next node
// every node distributed to three lists
while (head != null) {
next = head.next;
head.next = null;
if (head.val <= pivot) {
if (sH == null) {
sH = head;
sT = head;
} else {
sT.next = head;
sT = head;
}
}else {
if (bH == null) {
bH = head;
bT = head;
} else {
bT.next = head;
bT = head;
}
}
head = next;
}
if (sT != null) {
sT.next = bH;
}
return sH != null ? sH : bH;
}
}

打印两个链表的公共值

现有两个升序链表，且链表中均无重复元素。请设计一个高效的算法，打印两个链表的公共值部分。

给定两个链表的头指针headA和headB，请返回一个vector，元素为两个链表的公共部分。请保证返回数组的升序。两个链表的元素个数均小于等于500。保证一定有公共值

# 测试样例：
{1,2,3,4,5,6,7},{2,4,6,8,10}
返回：[2.4.6]

我的提交

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Common:
def findCommonParts(self, headA, headB):
# write code here
if not headA or not headB:
return []
result = []
while headA and headB:
if headA.val == headB.val:
result.append(headA.val)
headA = headA.next
headB = headB.next
elif headA.val < headB.val:
headA = headA.next
else:
headB = headB.next
return result

参考答案

import java.util.*;

/*
public class ListNode {
int val;
ListNode next = null;

ListNode(int val) {
this.val = val;
}
}*/
public class Common {
public int[] findCommonParts(ListNode head1, ListNode head2) {
LinkedList list = new LinkedList();
while (head1 != null && head2 != null) {
if (head1.val < head2.val) {
head1 = head1.next;
} else if (head1.val > head2.val) {
head2 = head2.next;
} else {
list.add(head1.val);
head1 = head1.next;
head2 = head2.next;
}
}
int[] res = new int[list.size()];
int index = 0;
while (!list.isEmpty()) {
res[index++] = list.pollFirst();
}
return res;
}
}