链表经典面试题
1 回文链表
1.1 判断方法
- 第一种（笔试）：
- 链表从中间分开，把后半部分的节点放到栈中
- 从链表的头结点开始，依次和弹出的节点比较
- 第二种（面试）：
- 反转链表的后半部分，中间节点的下一节点指向空
- 两个指针分别从链表的头和尾出发比较
- 直到左边节点到空或两个节点都到空停止
- 判断结果出来后，要把链表后半部分反转回去。
1.2 代码实现
ublic class IsPalindromList {
public static class Node {
public int val;
public Node next;
public Node(int data) {
val = data;
next = null;
}
}
public static boolean isPalindrom1(Node head) {
if (head == null || head.next == null) {
return true;
}
Stack<Node> stack = new Stack<>();
Node mid = head;
Node fast = head;
while (fast.next != null && fast.next.next != null) {
mid = mid.next;
fast = fast.next.next;
}
while (mid.next != null) {
mid = mid.next;
stack.add(mid);
}
Node i = head;
boolean ans = true;
while (stack.size() != 0) {
if (i.val != stack.pop().val) {
ans = false;
break;
}
i = i.next;
}
return ans;
}
public static boolean isPalindrom2(Node head) {
if (head == null || head.next == null) {
return true;
}
// 找到中间节点
Node mid = head;
Node fast = head;
while (fast.next != null && fast.next.next != null) {
mid = mid.next;
fast = fast.next.next;
}
Node cur = mid;
// 后边段链表反转
Node pre = null;
Node next = null;
while (cur != null) {
next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
Node left = head;
Node right = pre;
boolean ans = true;
while (left != null && right != null) {
if (left.val != right.val) {
ans = false;
break;
}
left = left.next;
right = right.next;
}
cur = pre;
pre = null;
next = null;
while (cur != null) {
next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
return ans;
}
public static void main(String[] args) {
Node head = new Node(0);
head.next = new Node(1);
head.next.next = new Node(2);
head.next.next.next = new Node(2);
head.next.next.next.next = new Node(1);
head.next.next.next.next.next = new Node(0);
System.out.println(isPalindrom2(head));
System.out.println(isPalindrom1(head));
}
}
2 荷兰国旗
链表的荷兰国旗问题，给定一个链表头节点，一个划分值
- 把链表中小于划分值的放在左边
- 等于划分值的放在中间
- 大于划分值的放在右边。
2.1 解决方法
（其实我第一次写的时候第二种方法比第一种方法写的快，因为思路简单，不绕。）
- 第一种（笔试）：
- 把链表的所有节点放在一个节点数组中
- 对这个数组做
partition过程 - 之后把数组每个节点连起来
- 第二种（面试）：
- 定义有六个节点，分别代表：大于、等于和小于区域的头和尾
- 链表开始遍历，比划分值小的连在小于区域下方，同时断开节点和之前链表的关系，指向空
- 等于和大于同理
- 遍历完链表之后，开始把小于区域的尾巴和等于区域的头连接，等于区域的尾巴和大于区域的头连接
2.2 代码实现
public class SmallEqualBig {
public static class Node {
public int val;
public Node next;
public Node(int val) {
this.val = val;
next = null;
}
}
public static Node listPartition1(Node head, int pivot) {
if (head == null || head.next == null) {
return head;
}
int index = 1;
Node cur = head;
while (cur.next != null) {
index++;
cur = cur.next;
}
Node[] arr = new Node[index];
cur = head;
for (int i = 0; i < arr.length; i++) {
arr[i] = cur;
cur = cur.next;
}
partition(arr,pivot);
for (index = 1; index != arr.length; index++) {
arr[index - 1].next = arr[index];
}
arr[index - 1].next = null;
return arr[0];
}
public static Node listPartition2(Node head, int pivot) {
if (head == null || head.next == null) {
return head;
}
Node smallHead = null;
Node smallTail = null;
Node equalHead = null;
Node equalTail = null;
Node bigHead = null;
Node bigTail = null;
Node next = null;
while (head != null) {
next = head.next;
head.next = null;
if (head.val < pivot) {
if (smallHead == null) {
smallHead = head;
smallTail = head;
}else {
smallTail.next = head;
smallTail = smallTail.next;
}
} else if (head.val == pivot) {
if (equalHead == null) {
equalHead = head;
equalTail = head;
}else {
equalTail.next = head;
equalTail = equalTail.next;
}
}else {
if (bigHead == null) {
bigHead = head;
bigTail = head;
}else {
bigTail.next = head;
bigTail = bigTail.next;
}
}
head = next;
}
if (smallTail != null) {
smallTail.next = equalHead;
equalTail = equalTail == null ? smallTail : equalTail;
}
if (equalTail != null) {
equalTail.next = bigHead;
}
return smallHead == null ? (equalHead == null ? bigHead : equalHead) : smallHead;
}
private static void partition(Node[] arr, int pivot) {
int small = -1;
int big = arr.length;
int index = 0;
while (index < big) {
if(arr[index].val < pivot) {
swap(arr, index++, ++small);
} else if (arr[index].val > pivot) {
swap(arr, index++, --big);
}else {
index++;
}
}
}
private static void swap(Node[] arr, int i, int j) {
Node temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static void main(String[] args) {
Node head = new Node(9);
head.next = new Node(1);
head.next.next = new Node(2);
head.next.next.next = new Node(6);
head.next.next.next.next = new Node(4);
head.next.next.next.next.next = new Node(5);
head.next.next.next.next.next.next = new Node(2);
Node node = listPartition1(head, 2);
while (node != null) {
System.out.print(node.val + " ");
node = node.next;
}
System.out.println();
Node head2 = new Node(9);
head2.next = new Node(1);
head2.next.next = new Node(2);
head2.next.next.next = new Node(6);
head2.next.next.next.next = new Node(4);
head2.next.next.next.next.next = new Node(5);
head2.next.next.next.next.next.next = new Node(2);
Node node1 = listPartition2(head2, 2);
while (node1 != null) {
System.out.print(node1.val + " ");
node1 = node1.next;
}
}
}
3 复制链表
复制特殊链表，单链表中加了个rand指针，可能指向任意一个节点，也可能指向null
给定这个链表的头节点，完成链表的复制
返回复制的新链表的头节点。时间复杂度O(N),额外空间复杂度O(1)
3.1 解决方法
-
方法一（笔试）：
- 用HashMap来解决，key是原数组节点，value是复制的数组节点
-
方式二（面试）：
-
将复制的新节点插入到原数组的节点中，如1–>2–>3,变成1–>(copy)1–>2–>(copy)2–>3–>(copy)3
-
复制节点的
rand指针指向就是原数组的节点的rand指针的下一节点（下图所示） -
把复制数组拿出来的时候注意复制数组连起来还要把原数组连起来
-
3.2 代码实现
public class CopyListWithRandom {
public static class Node {
public int val;
public Node next;
public Node rand;
public Node(int val) {
this.val = val;
}
}
// map方法
public static Node copyListWithRandom(Node head) {
if (head == null) {
return null;
}
HashMap<Node, Node> map = new HashMap<>();
Node cur = head;
while (cur != null) {
map.put(cur, new Node(cur.val));
cur = cur.next;
}
Node ans = map.get(head);
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 ans;
}
// 不借助容器
public static Node copyListWithRandom2(Node head) {
if (head == null) {
return null;
}
Node cur = head;
Node next = null;
while (cur != null) {
next = cur.next;
cur.next = new Node(cur.val);
cur.next.next = next;
cur = next;
}
Node copyNode = null;
cur = head;
while (cur != null) {
copyNode = cur.next;
copyNode.rand = cur.rand == null ? null : cur.rand.next;
cur = copyNode.next;
}
Node ans = head.next;
cur = head;
while (cur != null) {
next = cur.next.next;
copyNode = cur.next;
copyNode.next = next == null ? null : next.next;
cur.next = next;
cur = next;
}
return ans;
}
public static void main(String[] args) {
Node head = new Node(1);
Node node2 = new Node(2);
Node node3 = new Node(3);
head.next = node2;
node2.next = node3;
head.rand = node3;
node2.rand = head;
Node node = copyListWithRandom(head);
while (node != null) {
System.out.println(node.val + " ");
System.out.println(node == head);
node = node.next;
head = head.next;
}
}
}
4 链表相交
给定两个可能有环也可能无环的单链表，给定头节点1和头节点2.
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
时间复杂度O(N),额外空间复杂度O(1)
4.1 解决方法
- 先判断两个链表是否有环
- 两个都无环
- 如果两个链表的末尾节点相等，那么就是相交的，用指针先把长的链表走完两个链表的差值后，两个链表开始同时走，直到两个节点相等，就是相交的点
- 如果末尾节点不相等，那么两个链表不相交
- 一个有环一个无环：两个链表不会有相交点
- 两个都有环
- 判断两个链表进环的第一个节点是否是相同的
- 相同：相交，且相交点就在无环的部分中，参考2
- 不相同：从一个链表的入环节点开始找，如果找到了第二个链表的入环节点，那么就相交，返回两个入环节点任意一个，如果转一圈到第一个链表的入环节点还没有找到第二个入环节点，此时不相交。
4.2代码实现
public class FindFirstIntersectNode {
public static class Node {
public int val;
public Node next;
public Node(int val) {
this.val = val;
}
}
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;
}
private static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {
Node cur1 = head1;
Node cur2 = head2;
if (loop1 == loop2) {
int n = 0;
while (cur1.next != null) {
n++;
cur1 = cur1.next;
}
while (cur2.next != null) {
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 loop2;
}
cur1 = cur1.next;
}
return null;
}
}
private static Node noLoop(Node head1, Node head2) {
Node cur1 = head1;
Node cur2 = head2;
int n = 0;
while (cur1.next != null) {
n++;
cur1 = cur2.next;
}
while (cur2.next != null) {
n--;
cur2 = cur2.next;
}
if (cur1 != cur2) {
return null;
}
cur1 = n < 0 ? head2 : head1;
cur2 = cur1 == head2 ? head1 : head2;
n = Math.abs(n);
while (n != 0) {
n--;
cur1 = cur1.next;
}
while (cur1 != cur2) {
cur1 = cur1.next;
cur2 = cur2.next;
}
return cur1;
}
private static Node getLoopNode(Node head) {
if (head.next == null || head.next.next ==null) {
return null;
}
Node slow = head.next;
Node fast = head.next.next;
while (fast != slow) {
if (fast == null) {
return null;
}
slow = slow.next;
fast = fast.next.next;
}
fast = head;
while (fast != slow) {
fast = fast.next;
slow = slow.next;
}
return fast;
}
public static void main(String[] args) {
Node head = new Node(0);
head.next = new Node(1);
head.next.next = new Node(2);
head.next.next.next = new Node(3);
head.next.next.next.next = new Node(4);
head.next.next.next.next.next = head.next.next;
Node head1 = new Node(0);
head1.next = new Node(1);
head1.next.next = new Node(2);
head1.next.next.next = new Node(3);
head1.next.next.next.next = new Node(4);
head1.next.next.next.next.next = head.next.next.next;
Node intersectNode = getIntersectNode(head, head1);
System.out.println(intersectNode.val);
}
}