算法初级面试题03——打印链表公共部分、判断链表是否为回文、按值划分链表为小于等于大于、复制随机指针链表、两链表相交判断的一系列问题
接着前面的内容
这次主要讨论链表相关的题目
机试的时候怎么快怎么做,面试的时候要聊时间O(N),额外空间复杂度达到O(1)。
题目十
打印两个有序链表的公共部分
【题目】 给定两个有序链表的头指针head1和head2,打印两个链表的公共部分。
public class Code_10_PrintCommonPart { public static class Node { public int value; public Node next; public Node(int data) { this.value = data; } } public static void printCommonPart(Node head1, Node head2) { System.out.print("Common Part: "); while (head1 != null && head2 != null) { if (head1.value < head2.value) { head1 = head1.next; } else if (head1.value > head2.value) { head2 = head2.next; } else { System.out.print(head1.value + " "); head1 = head1.next; head2 = head2.next; } } System.out.println(); } public static void printLinkedList(Node node) { System.out.print("Linked List: "); while (node != null) { System.out.print(node.value + " "); node = node.next; } System.out.println(); } public static void main(String[] args) { Node node1 = new Node(2); node1.next = new Node(3); node1.next.next = new Node(5); node1.next.next.next = new Node(6); Node node2 = new Node(1); node2.next = new Node(2); node2.next.next = new Node(5); node2.next.next.next = new Node(7); node2.next.next.next.next = new Node(8); printLinkedList(node1); printLinkedList(node2); printCommonPart(node1, node2); } }
题目十一
判断一个链表是否为回文结构
【题目】 给定一个链表的头节点head,请判断该链表是否为回文结构。 例如: 1->2->1,返回true。 1->2->2->1,返回true。15->6->15,返回true。 1->2->3,返回false。
进阶: 如果链表长度为N,时间复杂度达到O(N),额外空间复杂度达到O(1)。
方法一:放到栈里面再逆序比对(利用额外空间)
方法二:两个指针,1走一步,2走两步,2到了终点,1就到了中点,然后1往后的压栈,在逐一比对。
方法三:(额外空间O(1)):先通过快慢指针,找到中点,然后改变中间后面的链表的指向,在逐一比对,最后再把指针还原如初。
public class Code_11_IsPalindromeList { public static class Node { public int value; public Node next; public Node(int data) { this.value = data; } } // need n extra space public static boolean isPalindrome1(Node head) { Stack<Node> stack = new Stack<Node>(); Node cur = head; while (cur != null) { stack.push(cur); cur = cur.next; } while (head != null) { if (head.value != stack.pop().value) { return false; } head = head.next; } return true; } // need n/2 extra space public static boolean isPalindrome2(Node head) { if (head == null || head.next == null) { return true; } //慢指针,提前走一步为了当链表为奇数时走在中间区域的最后一个数 //也方便了直接存入在对比的栈当中 Node slow = head.next; //快指针 Node fast = head; while (fast.next != null && fast.next.next != null) { slow = slow.next; fast = fast.next.next; } Stack<Node> stack = new Stack<Node>(); while (slow != null) { stack.push(slow); slow = slow.next; } while (!stack.isEmpty()) { if (head.value != stack.pop().value) { return false; } head = head.next; } return true; } // need O(1) extra space public static boolean isPalindrome3(Node head) { if (head == null || head.next == null) { return true; } Node n1 = head; Node n2 = head; while (n2.next != null && n2.next.next != null) { // find mid node n1 = n1.next; // n1 -> mid n2 = n2.next.next; // n2 -> end } n2 = n1.next; // n2 -> right part first node n1.next = null; // mid.next -> null Node n3 = null; while (n2 != null) { // right part convert n3 = n2.next; // n3 -> save next node 操作前先保存下一个节点 n2.next = n1; // next of right node convert 反转指向 n1 = n2; // n1 move 保存当前节点,作为后一个的前节点 n2 = n3; // n2 move 向前推进 } n3 = n1; // n3 -> save last node n2 = head;// n2 -> left first node boolean res = true; while (n1 != null && n2 != null) { // check palindrome if (n1.value != n2.value) { res = false; break; } n1 = n1.next; // left to mid n2 = n2.next; // right to mid } n1 = n3.next; n3.next = null; while (n1 != null) { // recover list n2 = n1.next; n1.next = n3; n3 = n1; n1 = n2; } return res; } public static void printLinkedList(Node node) { System.out.print("Linked List: "); while (node != null) { System.out.print(node.value + " "); node = node.next; } System.out.println(); } public static void main(String[] args) { Node head = null; printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(2); head.next.next.next = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(2); head.next.next.next.next = new Node(1); printLinkedList(head); System.out.print(isPalindrome1(head) + " | "); System.out.print(isPalindrome2(head) + " | "); System.out.println(isPalindrome3(head) + " | "); printLinkedList(head); System.out.println("========================="); } }
题目十二
将单向链表按某值划分成左边小、中间相等、右边大的形式
【题目】 给定一个单向链表的头节点head,节点的值类型是整型,再给定一个整 数pivot。实现一个调整链表的函数,将链表调整为左部分都是值小于 pivot 的节点,中间部分都是值等于pivot的节点,右部分都是值大于 pivot的节点。除这个要求外,对调整后的节点顺序没有更多的要求。 例如:链表9->0->4->5->1,pivot=3。 调整后链表可以是1->0->4->9->5,也可以是0->1->9->5->4。总之,满 足左部分都是小于3的节点,中间部分都是等于3的节点(本例中这个部分为空),右部分都是大于3的节点即可。对某部分内部的节点顺序不做 要求。
可以使用荷兰国旗的方法去处理,数组每个元素变为结点类型,然后再接起来。
进阶: 在原问题的要求之上再增加如下两个要求。
在左、中、右三个部分的内部也做顺序要求,要求每部分里的节点从左 到右的顺序与原链表中节点的先后次序一致。 例如:链表9->0->4->5->1,pivot=3。调整后的链表是0->1->9->4->5。 在满足原问题要求的同时,左部分节点从左到右为0、1。在原链表中也 是先出现0,后出现1;中间部分在本例中为空,不再讨论;右部分节点 从左到右为9、4、5。在原链表中也是先出现9,然后出现4,最后出现5。
如果链表长度为N,时间复杂度请达到O(N),额外空间复杂度请达到O(1)。
准备三个变量,这三个变量都是节点对象的引用类型。less eq more
先遍历链表,找到第一个小于/等于/大于num的节点,让less/eq/more等于那个节点。
然后准备多一个end,每次加入一个,end就加一,直到最后把三个小链表头尾链接起来。
(原理是把一个大链表,拆成三个小链表,再组装起来)有限几个变量O(1)。
public class Code_12_SmallerEqualBigger { public static class Node { public int value; public Node next; public Node(int data) { this.value = data; } } public static Node listPartition1(Node head, int pivot) { if (head == null) { return head; } Node cur = head; int i = 0; while (cur != null) { i++; cur = cur.next; } Node[] nodeArr = new Node[i]; i = 0; cur = head; for (i = 0; i != nodeArr.length; i++) { nodeArr[i] = cur; cur = cur.next; } arrPartition(nodeArr, pivot); for (i = 1; i != nodeArr.length; i++) { nodeArr[i - 1].next = nodeArr[i]; } nodeArr[i - 1].next = null; return nodeArr[0]; } public static void arrPartition(Node[] nodeArr, int pivot) { int small = -1; int big = nodeArr.length; int index = 0; while (index != big) { if (nodeArr[index].value < pivot) { swap(nodeArr, ++small, index++); } else if (nodeArr[index].value == pivot) { index++; } else { swap(nodeArr, --big, index); } } } public static void swap(Node[] nodeArr, int a, int b) { Node tmp = nodeArr[a]; nodeArr[a] = nodeArr[b]; nodeArr[b] = tmp; } public static Node listPartition2(Node head, int pivot) { Node sH = null; // small head Node sT = null; // small tail Node eH = null; // equal head Node eT = null; // equal tail Node bH = null; // big head Node bT = null; // big tail Node next = null; // save next node // every node distributed to three lists while (head != null) { next = head.next; head.next = null; if (head.value < pivot) { if (sH == null) { sH = head; sT = head; } else { sT.next = head; sT = head; } } else if (head.value == pivot) { if (eH == null) { eH = head; eT = head; } else { eT.next = head; eT = head; } } else { if (bH == null) { bH = head; bT = head; } else { bT.next = head; bT = head; } } head = next; } // small and equal reconnect if (sT != null) { sT.next = eH; eT = eT == null ? sT : eT; } // all reconnect if (eT != null) { eT.next = bH; } return sH != null ? sH : eH != null ? eH : bH; } public static void printLinkedList(Node node) { System.out.print("Linked List: "); while (node != null) { System.out.print(node.value + " "); node = node.next; } System.out.println(); } public static void main(String[] args) { Node head1 = new Node(7); head1.next = new Node(9); head1.next.next = new Node(1); head1.next.next.next = new Node(8); head1.next.next.next.next = new Node(5); head1.next.next.next.next.next = new Node(2); head1.next.next.next.next.next.next = new Node(5); printLinkedList(head1); // head1 = listPartition1(head1, 4); head1 = listPartition2(head1, 5); printLinkedList(head1); } }
题目十三
复制含有随机指针节点的链表
【题目】 一种特殊的链表节点类描述如下:
public class Node { public int value; public Node next; public Node rand; public Node(int data) {this.value = data;} }
Node类中的value是节点值,next指针和正常单链表中next指针的意义一 样,都指向下一个节点,rand指针是Node类中新增的指针,这个指针可 能指向链表中的任意一个节点,也可能指向null。 给定一个由Node节点类型组成的无环单链表的头节点head,请实现一个 函数完成这个链表中所有结构的复制,并返回复制的新链表的头节点。
进阶:不使用额外的数据结构,只用有限几个变量,且在时间复杂度为 O(N) 内完成原问题要实现的函数。
方法一:准备一个hashmap,依次把结点作为key,复制的结点为value存入。然后再通过key查找value的方式,复制指针的指向,达到深入拷贝。
进阶方法:不用hashmap的方法,先遍历一遍原链表,形成1-->1’ --->2...的结构,然后利用1-->random指引1’-->random的连接,连完再分离。
public class Code_13_CopyListWithRandom { public static class Node { public int value; public Node next; public Node rand; public Node(int data) { this.value = data; } } public static Node copyListWithRand1(Node head) { HashMap<Node, Node> map = new HashMap<Node, Node>(); Node cur = head; while (cur != null) { map.put(cur, new Node(cur.value)); cur = cur.next; } cur = head; while (cur != null) { map.get(cur).next = map.get(cur.next); map.get(cur).rand = map.get(cur.rand); cur = cur.next; } return map.get(head); } public static Node copyListWithRand2(Node head) { if (head == null) { return null; } Node cur = head; Node next = null; // copy node and link to every node //先遍历一遍原链表,形成1-->1’--->2...的结构 while (cur != null) { next = cur.next; cur.next = new Node(cur.value); cur.next.next = next; cur = next; } cur = head; Node curCopy = null; // set copy node rand //复制随机指针 while (cur != null) { next = cur.next.next; curCopy = cur.next; ////存在一种情况,末尾的随机指针指向为空,再引用空的下一个就会报错 curCopy.rand = cur.rand != null ? cur.rand.next : null; cur = next; } Node res = head.next; cur = head; // split //分离的同时,顺便重连链表 while (cur != null) { next = cur.next.next;//2 null curCopy = cur.next;//1' 3' cur.next = next;//2 null //需要判断后续是否还有节点的存在 curCopy.next = next != null ? next.next : null;//2' cur = next;//2 } return res; } public static void printRandLinkedList(Node head) { Node cur = head; System.out.print("order: "); while (cur != null) { System.out.print(cur.value + " "); cur = cur.next; } System.out.println(); cur = head; System.out.print("rand: "); while (cur != null) { System.out.print(cur.rand == null ? "- " : cur.rand.value + " "); cur = cur.next; } System.out.println(); } public static void main(String[] args) { Node head = null; Node res1 = null; Node res2 = null; printRandLinkedList(head); res1 = copyListWithRand1(head); printRandLinkedList(res1); res2 = copyListWithRand2(head); printRandLinkedList(res2); printRandLinkedList(head); System.out.println("========================="); head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); head.next.next.next.next.next = new Node(6); head.rand = head.next.next.next.next.next; // 1 -> 6 head.next.rand = head.next.next.next.next.next; // 2 -> 6 head.next.next.rand = head.next.next.next.next; // 3 -> 5 head.next.next.next.rand = head.next.next; // 4 -> 3 head.next.next.next.next.rand = null; // 5 -> null head.next.next.next.next.next.rand = head.next.next.next; // 6 -> 4 printRandLinkedList(head); res1 = copyListWithRand1(head); printRandLinkedList(res1); res2 = copyListWithRand2(head); printRandLinkedList(res2); printRandLinkedList(head); System.out.println("========================="); } }
题目十四
两个单链表相交的一系列问题
【题目】 在本题中,单链表可能有环,也可能无环。给定两个单链表的头节点 head1和head2,这两个链表可能相交,也可能不相交。请实现一个函数, 如果两个链表相交,请返回相交的第一个节点;如果不相交,返回null 即可。
要求:如果链表1的长度为N,链表2的长度为M,时间复杂度请达到 O(N+M),额外空间复杂度请达到O(1)。
首先解决判断是否有环,利用hashset来做(有环就返回第一个入环的结点)
public static Node getLoopNodeByHashMap(Node head){ if(head == null || head.next == null||head.next.next==null ){ return null; } HashSet<Node> nodeset = new HashSet<>(); while (head!=null){ if(nodeset.contains(head)){ return head; } nodeset.add(head); head = head.next; } return null; }
不用hash表要怎么做?准备两个指针,一快一慢,快一次两步,慢一次一步,如果快指针走到null直接返回无环。
如果快指针和慢指针相遇,证明有环,相遇后,快指针回到起点,快指针从一次两步变一次一步,快指针和慢指针一定在第一个入环结点处相遇。(数学归纳法)(玄学)
public static Node getLoopNode(Node head) { if (head == null || head.next == null || head.next.next == null) { return null; } Node n1 = head.next; // n1 -> slow Node n2 = head.next.next; // n2 -> fast while (n1 != n2) { if (n2.next == null || n2.next.next == null) { return null; } n2 = n2.next.next; n1 = n1.next; } n2 = head; // n2 -> walk again from head while (n1 != n2) { n1 = n1.next; n2 = n2.next; } return n1; }
判断链表是否相交:
使用map查看两个无环链表是否相交,先链表1放入map,然后遍历链表2,第一个出现在map中的就是相交的节点,否则没有就不相交。
Loop是第一个入环的结点。
public static Node noLoopByHashSet(Node head1, Node head2) { if (head1 == null || head2 == null) { return null; } HashSet<Node> NodeSet = new HashSet<>(); while (head1 != null) { NodeSet.add(head1); head1 = head1.next; } while (head2 != null) { if (NodeSet.contains(head2)) { return head2; } head2 = head2.next; } return null; }
不用Map怎么做?遍历链表1统计长度,并拿到链表1最后一个节点。链表2同上操作。
接着判断end1和end2内存地址是否一致,如果不相等,他们不可能相交。
如果相等,证明相交,但是不证明这是他们第一个相交的节点。
当相等时候,进行利用长度进行后序的操作
例如:1长为100,2为80,那么1先走20步
然后和2一起走,他们肯定能走到第一个相会处。
public static Node noLoop(Node head1, Node head2) { if (head1 == null || head2 == null) { return null; } Node cur1 = head1; Node cur2 = head2; int n = 0; while (cur1.next != null) { n++; cur1 = cur1.next; } while (cur2.next != null) { n--; cur2 = cur2.next; } if (cur1 != cur2) { return null; } cur1 = n > 0 ? head1 : head2;//找出哪个比较长 cur2 = cur1 == head1 ? head2 : head1; n = Math.abs(n); while (n != 0) {//较长的先走 n--; cur1 = cur1.next; } while (cur1 != cur2) { cur1 = cur1.next; cur2 = cur2.next; } return cur1; }
如果一个有环一个无环?结论:不可能相交
两个有环链表相交,怎么找到第一个入环的节点?
有三种拓扑结构,
当loop1==loop2(loop是第一个入环的节点)时候,是第二种(可以复用无环链表相交问题的处理逻辑)
如果不相等可能是结构一或者三,怎么区分?第一个loop1一直next如果一直这样都转回自己了,还没遇到loop2就是第一种拓扑。(不相交,返回空)
如果他遇到了loop2,就是第三种拓扑。此时返回loop1/2作为相交的节点都对。
public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) { Node cur1 = null; Node cur2 = null; if (loop1 == loop2) { cur1 = head1; cur2 = head2; int n = 0; while (cur1 != loop1) { n++; cur1 = cur1.next; } while (cur2 != loop2) { n--; cur2 = cur2.next; } cur1 = n > 0 ? head1 : head2; cur2 = cur1 == head1 ? head2 : head1; n = Math.abs(n); while (n != 0) { n--; cur1 = cur1.next; } while (cur1 != cur2) { cur1 = cur1.next; cur2 = cur2.next; } return cur1; } else { cur1 = loop1.next; while (cur1 != loop1) { if (cur1 == loop2) { return loop1; } cur1 = cur1.next; } return null; } }
两链表相交的一系列问题全部代码(包含测试代码)
public class Code_14_FindFirstIntersectNode { public static class Node { public int value; public Node next; public Node(int data) { this.value = data; } } public static Node getIntersectNode(Node head1, Node head2) { if (head1 == null || head2 == null) { return null; } Node loop1 = getLoopNode(head1); Node loop2 = getLoopNode(head2); if (loop1 == null && loop2 == null) { return noLoop(head1, head2); } if (loop1 != null && loop2 != null) { return bothLoop(head1, loop1, head2, loop2); } return null; } public static Node getLoopNodeByHashMap(Node head){ if(head == null || head.next == null||head.next.next==null ){ return null; } HashSet<Node> nodeset = new HashSet<>(); while (head!=null){ if(nodeset.contains(head)){ return head; } nodeset.add(head); head = head.next; } return null; } public static Node getLoopNode(Node head) { if (head == null || head.next == null || head.next.next == null) { return null; } Node n1 = head.next; // n1 -> slow Node n2 = head.next.next; // n2 -> fast while (n1 != n2) { if (n2.next == null || n2.next.next == null) { return null; } n2 = n2.next.next; n1 = n1.next; } n2 = head; // n2 -> walk again from head while (n1 != n2) { n1 = n1.next; n2 = n2.next; } return n1; } public static Node noLoop(Node head1, Node head2) { if (head1 == null || head2 == null) { return null; } Node cur1 = head1; Node cur2 = head2; int n = 0; while (cur1.next != null) { n++; cur1 = cur1.next; } while (cur2.next != null) { n--; cur2 = cur2.next; } if (cur1 != cur2) { return null; } cur1 = n > 0 ? head1 : head2;//找出哪个比较长 cur2 = cur1 == head1 ? head2 : head1; n = Math.abs(n); while (n != 0) {//较长的先走 n--; cur1 = cur1.next; } while (cur1 != cur2) { cur1 = cur1.next; cur2 = cur2.next; } return cur1; } public static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) { Node cur1 = null; Node cur2 = null; if (loop1 == loop2) { cur1 = head1; cur2 = head2; int n = 0; while (cur1 != loop1) { n++; cur1 = cur1.next; } while (cur2 != loop2) { n--; cur2 = cur2.next; } cur1 = n > 0 ? head1 : head2; cur2 = cur1 == head1 ? head2 : head1; n = Math.abs(n); while (n != 0) { n--; cur1 = cur1.next; } while (cur1 != cur2) { cur1 = cur1.next; cur2 = cur2.next; } return cur1; } else { cur1 = loop1.next; while (cur1 != loop1) { if (cur1 == loop2) { return loop1; } cur1 = cur1.next; } return null; } } public static void main(String[] args) { // 1->2->3->4->5->6->7->null Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(4); head1.next.next.next.next = new Node(5); head1.next.next.next.next.next = new Node(6); head1.next.next.next.next.next.next = new Node(7); // 0->9->8->6->7->null Node head2 = new Node(0); head2.next = new Node(9); head2.next.next = new Node(8); head2.next.next.next = head1.next.next.next.next.next; // 8->6 System.out.println(getIntersectNode(head1, head2).value); // 1->2->3->4->5->6->7->4... head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(4); head1.next.next.next.next = new Node(5); head1.next.next.next.next.next = new Node(6); head1.next.next.next.next.next.next = new Node(7); head1.next.next.next.next.next.next = head1.next.next.next; // 7->4 // 0->9->8->2... head2 = new Node(0); head2.next = new Node(9); head2.next.next = new Node(8); head2.next.next.next = head1.next; // 8->2 System.out.println(getIntersectNode(head1, head2).value); // 0->9->8->6->4->5->6.. head2 = new Node(0); head2.next = new Node(9); head2.next.next = new Node(8); head2.next.next.next = head1.next.next.next.next.next; // 8->6 System.out.println(getIntersectNode(head1, head2).value); } }
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