郑州网站建设 易云互联东莞营销

目录

一、分隔链表

二、旋转链表

三、删除链表中重复的数字

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一、分隔链表

用两个list存一下小于和大于等于 x的节点

最后串起来就行

 public ListNode partition(ListNode head, int x) {ListNode ret = new ListNode(1);ListNode cur = ret;List<ListNode> small = new ArrayList<>();List<ListNode> big = new ArrayList<>();List<ListNode> equal = new ArrayList<>();while(head != null){if(head.val < x){small.add(head);} else if(head.val >= x){big.add(head);}head = head.next;}for(int i = 0;i < small.size();i++){cur.next = small.get(i);cur = cur.next;}for(int i = 0;i < big.size();i++){cur.next = big.get(i);cur = cur.next;}if(big.size() > 0){big.get(big.size() - 1).next = null;} else if(small.size() > 0){small.get(small.size() - 1).next = null;}return ret.next;    }

二、旋转链表

题目意思就是每次把末尾最后一个放到最左边

可以用list集合存一下 每次去掉最后一个，并把对其头插

需要考虑一下k特别大的情况  取余一下

 public ListNode rotateRight(ListNode head, int k) {List<ListNode> list = new LinkedList<>();ListNode ret = new ListNode(1);ListNode cur = ret;while(head != null){list.add(head);head = head.next;}if(list.size() > 0){k = k % list.size();}while(k-- != 0){if(list.size() > 0){ListNode first = list.get(list.size() - 1);list.remove(list.size() - 1);list.addFirst(first);}}for(int i = 0;i < list.size();i++){cur.next = list.get(i);cur = cur.next;}cur.next = null;return ret.next;}

三、删除链表中重复的数字

用map记录一下，哪些是重复数字，哪些不是重复数字

public ListNode deleteDuplicates(ListNode head) {Map<Integer,Boolean> map = new HashMap<>();ListNode cur = head;while(cur != null){if(map.containsKey(cur.val)){map.put(cur.val,false);}else{map.put(cur.val,true);}cur = cur.next;}cur = head;ListNode ret = new ListNode(1);ListNode curRet = ret;while(cur != null){if(map.get(cur.val)){curRet.next = cur;curRet = curRet.next;}cur = cur.next;}curRet.next = null;return ret.next;}