【LeetCode】Remove Duplicates from Sorted List I II

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I、每个数据只允许出现1次
     Remove Duplicates from Sorted List 
     Total Accepted: 7120 Total Submissions: 20880 My Submissions
     Given a sorted linked list, delete all duplicates such that each element appear only once.
     For example,
     Given 1->1->2, return 1->2.
     Given 1->1->2->3->3, return 1->2->3.
II、每个数据最多允许出现2次，注意这里是留下仅出现一次的数据
      Remove Duplicates from Sorted List II 
      Total Accepted: 4962 Total Submissions: 20108 My Submissions
      Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
      For example,
      Given 1->2->3->3->4->4->5, return 1->2->5.
      Given 1->1->1->2->3, return 2->3.
         其实和LeetCode/Remove Duplicates from Sorted Array I && II 一样，不同的只是，一个是数组，一个是链表。
         处理过程也基本类似。
         I、声明初始数据，int start = head.val; 然后循环head，依次比较数组，如果和start不相等，就把当前的val赋值给另外一个链表，
         II、这个也是判断head.val出现的次数。
        1、使用了map，扫描一遍链表，统计每个数字出现的次数。这里没有使用array[head.val]++，是因为head.val有可能为负值。第二次扫描的时候，判断当前的head.val出现了几次，如果是一次，就赋值给新链表。
         2、延续了Remove Duplicates from Sorted Array II的办法。有点复杂，而且速度也没有第一种快，也难理解。仍然是int start = head.val，然后每次扫描的时候，判断head.val和start是否相等，如果不相等，需要判断start出现几次。
         其实也就是每次给新链表赋值的时候，都是start。建议理解第一种方法。
I、Java AC

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {public ListNode deleteDuplicates(ListNode head) {if(head == null){return null;}ListNode node = new ListNode(head.val);ListNode point = node;int start = head.val;head = head.next;while(head != null){if(head.val != start){start = head.val;point.next = new ListNode(head.val);point = point.next;}head = head.next;}return node;}
}

II、Java AC(1,建议采纳)

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {public ListNode deleteDuplicates(ListNode head) {if(head == null){return null;}Map<Integer,Integer> numMap = new HashMap<Integer,Integer>();ListNode point = head;while(point != null){int val = point.val;int num = 1;if(numMap.containsKey(val)){num += numMap.get(val);}numMap.put(val,num);point = point.next;}ListNode node = new ListNode(0);point = node;while(head != null){int val = head.val;int num = numMap.get(val);if(num == 1){point.next = new ListNode(val);point = point.next;}head = head.next;}return node.next;}
}

II、Java AC(2,难理解)

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {public ListNode deleteDuplicates(ListNode head) {if(head == null){return null;}Map<Integer,Integer> numMap = new HashMap<Integer,Integer>();ListNode point = head;while(point != null){int val = point.val;int num = 1;if(numMap.containsKey(val)){num += numMap.get(val);}numMap.put(val,num);point = point.next;}ListNode node = new ListNode(0);point = node;while(head != null){int val = head.val;int num = numMap.get(val);if(num == 1){point.next = new ListNode(val);point = point.next;}head = head.next;}return node.next;}
}

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