JavaArithmetic/src/LeetCode/LeetCode203.java at master · freestylewill/JavaArithmetic · GitHub

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package LeetCode;

public class LeetCode203 {

// https://leetcode.com/problems/remove-linked-list-elements/description/

// 不使用虚拟头结点

// 时间复杂度: O(n)

// 空间复杂度: O(1)

public ListNode removeElements(ListNode head, int val) {

// 需要对头结点进行特殊处理

while(head != null && head.val == val){

ListNode node = head;

head = head.next;

}

if(head == null)

return head;

ListNode cur = head;

while(cur.next != null){

if(cur.next.val == val){

ListNode delNode = cur.next;

cur.next = delNode.next;

}

else

cur = cur.next;

}

return head;

}

// 弄了一个虚拟节点(不用对头结点再做处理了)

public ListNode removeElements2(ListNode head, int val) {

// 创建虚拟头结点

ListNode dummyHead = new ListNode(0);

dummyHead.next = head;

ListNode cur = dummyHead;

while(cur.next != null){

if(cur.next.val == val ){

ListNode delNode = cur.next;

cur.next = delNode.next;

}

else

cur = cur.next;

}

return dummyHead.next;

}

// 递归的方式来做

public ListNode removeElements3(ListNode head, int val) {

if(head == null)

return head;

// 用下一个节点跟value比较

ListNode res = removeElements3(head.next, val);

if(head.val == val)

return res;

else{

head.next = res;

return head;

}

// 递归的方式来做--->其实就是一个子串删除的问题。

public ListNode removeElements4(ListNode head, int val) {

if(head == null)

return head;

// 删除重复的节点，拿到子串

head.next = removeElements4(head.next, val);

// 如果头节点是要删除的节点，返回子串。如果不是，返回当前子串

return head.val == val ? head.next : head;

}

public static void main(String[] args) {

int[] nums = {1, 2, 6, 3, 4, 5, 6};

ListNode head = new ListNode(nums);

System.out.println(head);

ListNode res = (new LeetCode203()).removeElements3(head, 6);

System.out.println(res);

}

JavaArithmetic/src/LeetCode/LeetCode203.java at master · freestylewill/JavaArithmetic · GitHub

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