版权声明: https://blog.csdn.net/qq_38386316/article/details/82960169
题目描述
删除链表中等于给定值val的所有节点。
示例:
输入:
输出:
思路 * 1:
申请一个头结点cur,并使用一个变量dummpy指向其初始的位置,cur 和 head同时向前遍历,head总比cur要快一个结点。当head中结点的值达到要求时(即head.val = val),使较慢的cur.next直接指向head的下一个结点(当然也可以是null),此时cur不用继续向前,head向前一步后,通过达到的结点后,cur继续向前。最终返回dummpy.next即可。
代码 * 1:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeElements(ListNode head, int val) {
ListNode cur = new ListNode(-1);
ListNode dummpy = cur;
while(head != null) {
if(head.val == val) {
cur.next = head.next;
}else {
cur.next = head;
cur = cur.next;
}
head = head.next;
}
return dummpy.next;
}
}
复杂度分析
- 时间复杂度:
- 空间复杂度:
思路 * 2
还是可以将上面的思路改成递归的思路,很是简便的代码量(但是数据量过大时,会造成栈的溢出)。
代码 * 2
if (head == null) return null;
head.next = removeElements(head.next, val);
return head.val == val ? head.next : head;
复杂度分析
- 时间复杂度:
- 空间复杂度:
完整代码:
package leetcode203;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
/**
* Created by 张超帅 on 2018/10/7.
*/
class ListNode {
int val;
ListNode next;
ListNode( int val) {this.val = val;}
}
class Solution {
public ListNode removeElements(ListNode head, int val) {
/*ListNode cur = new ListNode(-1);
ListNode dummpy = cur;
while(head != null) {
if(head.val == val) {
cur.next = head.next;
}else {
cur.next = head;
cur = cur.next;
}
head = head.next;
}
return dummpy.next;
*/
if (head == null) return null;
head.next = removeElements(head.next, val);
return head.val == val ? head.next : head;
}
}
public class leetcode203 {
public static int[] stringToArrays(String input) {
input = input.trim();
input = input.substring(1,input.length() - 1);
if(input == null){
return new int[0];
}
String [] parts = input.split(",");
int[] res = new int[parts.length];
for(int i = 0; i < parts.length; i ++) {
res[i] = Integer.parseInt(parts[i].trim());
}
return res;
}
public static ListNode stringToListNode(String input) {
int[] nodes = stringToArrays(input);
ListNode cur = new ListNode(-1);
ListNode dummpy = cur; // 不能写成 ListNode dummpy = cur.next;
for(int node : nodes) {
cur.next = new ListNode(node);
cur = cur.next;
}
System.out.println(dummpy.next.val);
return dummpy.next;
}
public static String listnodeTostring(ListNode head) {
if(head == null) return "[]";
String res = "";
while(head != null) {
res += head.val + ", ";
head = head.next;
}
return "[" + res.substring(0, res.length() - 2) + "]";
}
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader( new InputStreamReader(System.in));
String line ;
while((line = in.readLine()) != null) {
ListNode head = stringToListNode(line);
int num = Integer.parseInt(in.readLine());
ListNode ret = new Solution().removeElements(head, num);
String result = listnodeTostring(ret);
System.out.println(result);
}
}
}