LeetCode Remove Duplicates from Sorted List & Remove Duplicates from Sorted List II

Remove Duplicates from Sorted List

Description:

Given a sorted linked list, delete all duplicates such that each element appear only once.

Solution:

每次while循环，都判断当前节点与下一个节点的val是否一样，一样则改变当前节点的next指针，否则跳到下一个节点。

import java.util.*;

public class Solution {
	public ListNode deleteDuplicates(ListNode head) {
		ListNode temp = head;
		ListNode next;
		while (temp != null) {
			next = temp.next;
			if (next == null)
				break;
			if (temp.val == next.val) {
				temp.next = next.next;
			} else {
				temp = next;
			}
		}
		return head;
	}
}
           

Remove Duplicates from Sorted List II

  Description:

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

Solution:

利用two pointer方法，每次对于当下要遍历的节点t1，都用进行另外一次遍历，记为t2，一直搜索到null或者t1和t2的val不同。如果t2是t1的next，那么就表示t1没有重复值，否则t1赋值到t2，继续遍历。

import java.util.*;

public class Solution {
	public ListNode deleteDuplicates(ListNode head) {
		ListNode t1, t2;

		t1 = head;
		t2 = head;
		while (t1 != null) {
			t2 = t1;
			while (t2 != null) {
				if (t2.val != t1.val)
					break;
				t2 = t2.next;
			}
			if (t1.next == t2) {
				break;
			} else {
				t1 = t2;
			}
		}

		head = t1;
		t1 = t2 = head;
		ListNode temp = head;
		while (t1 != null) {
			t2 = t1;
			while (t2 != null) {
				if (t2.val != t1.val)
					break;
				t2 = t2.next;
			}
			if (t1.next == t2) {
				temp.next = t1;
				temp = temp.next;
				System.out.println(temp.val);
				t1.next = null;
				t1 = t2;
			} else {
				t1 = t2;
			}
		}

		return head;
	}
}