• [LeetCode]Remove Duplicates from Sorted Array


    题目:Remove Duplicates from Sorted Array

    删除数组中重复元素。

    public class Solution {
        public int removeDuplicates(int[] nums) {
            int i = 0,j = 1;
            while(j < nums.length){
                if(nums[j] != nums[j - 1]){
                    nums[i++] = nums[j - 1];
                    if(j == nums.length - 1)nums[i++] = nums[j];
                }else if(j == nums.length - 1){
                    nums[i++] = nums[j];
                }
                j++;
            }
            if(nums.length == 1)return 1;
            return i;
        }
    }

    题目:Remove Duplicates from Sorted ArrayII

    保证数组中同一元素重复次数不超过2次。

    public class Solution {
        public int removeDuplicates(int[] nums) {
            if(nums.length == 1)return 1;
            int count = 0,i = 0,j = 1;
            while(j < nums.length){
                if(nums[j - 1] == nums[j]){
                    count++;
                }else{
                    if(count >= 1){
                        nums[i++] = nums[j - 2];
                        count = 0;
                    }
                    nums[i++] = nums[j - 1];
                }
                if(j == nums.length - 1){
                    if(nums[j - 1] == nums[j])nums[i++] = nums[j - 1];
                    nums[i++] = nums[j];
                }
                j++;
            }
            return i;
        }
    }

    题目:Remove Duplicates from Sorted List II

    删除已序链表中的重复元素。

    package com.example.medium;
    
    /**
     * 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.
     * @author FuPing
     *
     */
    public class DeleteDuplicates {
        public class ListNode {
            int val;
            ListNode next;
            ListNode(int x) { val = x; }
        }
        public ListNode deleteDuplicates(ListNode head) {
            if(head == null || head.next == null)return head;//头为空
            ListNode p = head.next,pre = head;
            boolean flag = false,headFlag = false;
            if(pre.val == p.val)headFlag = true;//头元素是否重复
            while(p.next != null){
                if(p.val == p.next.val){
                    p.next = p.next.next;
                    flag = true;
                }else{
                    if(flag){//有重复元素
                        pre.next = p.next;
                        flag = false;
                    }else{
                        pre = pre.next;
                    }
                    p = p.next;
                }
            }
            if(flag)pre.next = p.next;
            if(headFlag){//删除头部
                if(head.next != null && head.val == head.next.val)
                    return head.next.next;
                return head.next;
            }
            return head;
        }
    
        public static void main(String[] args) {
            // TODO Auto-generated method stub
    
        }
    
    }
  • 相关阅读:
    【BZOJ 1185】 凸包+旋转卡壳
    【BZOJ 2829】 2829: 信用卡凸包 (凸包)
    【BZOJ 1045】 1045: [HAOI2008] 糖果传递
    【BZOJ 2453|bzoj 2120】 2453: 维护队列 (分块+二分)
    【BZOJ 3343 】 分块
    【BZOJ 1069】 凸包+旋转卡壳
    【NOIP 2016 总结】
    【无聊放个模板系列】洛谷 负环 模板
    【无聊放个模板系列】BZOJ 3172 (AC自动机)
    【无聊放个模板系列】HDU 3506 (四边形不等式优化DP-经典石子合并问题[环形])
  • 原文地址:https://www.cnblogs.com/yeqluofwupheng/p/6685537.html
Copyright © 2020-2023  润新知