作业帮：最长连续序列(头部插入)

题目描述　

给定一个未排序的整数数组，找出最长连续序列的长度。

要求算法的时间复杂度为 O(n)。

示例:

输入: [100, 4, 200, 1, 3, 2]
输出: 4
解释: 最长连续序列是 [1, 2, 3, 4]。它的长度为 4。

来源：力扣（LeetCode） 128
链接：https://leetcode-cn.com/problems/longest-consecutive-sequence
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

代码1：

import java.util.HashMap;
import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine();
        String[] vals= str.substring(1, str.length()-1).split(",");
        long[] num = new long[vals.length];
        for (int i = 0; i < num.length; i++) {
            num[i] = Long.valueOf(vals[i].trim());
        }
        System.out.println(getLongestLegth(num));
    }
    public static long getLongestLegth(long[] num){
        long max = 0;
        HashMap<Long,Long> map = new HashMap<Long,Long>();
        for (int i = 0; i < num.length; i++) {
            if(! map.containsKey(num[i])){
                long left = map.get(num[i]-1)==null? 0: map.get(num[i]-1);//找到当前数的左边连续长度left
                long right = map.get(num[i]+1)==null? 0: map.get(num[i]+1);//找到当前述的右边连续长度right
                long current = 1 + left + right;//连续最大长度
                if(current > max){
                    max = current;//更新最大长度
                }
                map.put(num[i], current);
                map.put(num[i]-left, current);//更新左边界最大长度
                map.put(num[i]+right, current);//更新右边界最大长度
            }
        }
        return max;
    }
}

代码2：

import java.util.HashSet;
import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine();
        String[] vals= str.substring(1, str.length()-1).split(",");
        int[] num = new int[vals.length];
        for (int i = 0; i < num.length; i++) {
            num[i] = Integer.valueOf(vals[i].trim());
        }
        System.out.println(getIntegerestLegth(num));
    }
    public static int getIntegerestLegth(int[] num){
        int max = 0;
        if(num.length<2){
            return num.length;
        }
        HashSet<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < num.length; i++) {
            set.add(num[i]);
        }
        for (int i = 0; i < num.length; i++) {
            if(set.remove(num[i])){
                int currentLongest = 1;
                int current = num[i];
                while(set.remove(current-1)){
                    current --;
                }
                currentLongest += (num[i]-current);
                current = num[i];
                while(set.remove(current+1)){
                    current ++;
                }
                currentLongest += (current-num[i]);
                max = Math.max(max, currentLongest);
            }
        }
        return max;
    }
}

代码3

import java.util.Arrays;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine();
        String[] vals= str.substring(1, str.length()-1).split(",");
        int[] num = new int[vals.length];
        for (int i = 0; i < num.length; i++) {
            num[i] = Integer.valueOf(vals[i].trim());
        }
        System.out.println(getIntegerestLegth(num));
    }
    public static int getIntegerestLegth(int[] num){
        int max = 0;
        if(num.length<2){
            return num.length;
        }
        Arrays.sort(num);
        PriorityQueue<Integer> maxQueue = new PriorityQueue<Integer>((a,b)->b-a);
        int currentLongest = 1;
        for (int i = 0; i < num.length -1; i++) {
            if(num[i+1] != num[i]){
                if(num[i+1] - num[i] ==1){
                    currentLongest++;
                }else{
                    currentLongest = 1;
                }
            }
            maxQueue.offer(currentLongest);
        }
        return maxQueue.peek();
    }
}