leetcode之longest-consecutive-sequence

問題描述：
Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

For example,
Given[100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is[1, 2, 3, 4]. Return its length:4.

Your algorithm should run in O(n) complexity.

題目要求演算法時間複雜度是O（n），所以先排序的思想是行不通的，這裡為了做比較，也實現了先排序的方法。
另一種演算法的思想就是用空間換時間：
採用hash表，先初始化一個hash表， 儲存所有數組元素， 然後遍曆這個數組， 對找到的數組元素， 去搜尋其相連的上下兩個元素是否在hash表中， 如果在， 刪除相應元素並增加此次尋找的資料長度， 如果不在， 從下一個元素出發尋找。

package suda.alex.leetcode;import java.util.Arrays;import java.util.HashSet;import java.util.Scanner;import java.util.Set;public class LongestConsecutive {    /**     * @param args     */    public static void main(String[] args) {        Scanner scanner = new Scanner(System.in);        System.out.println("input the size of num:");        int len = scanner.nextInt();        System.out.println("input num:");        int[] num = new int[len];        for (int i = 0; i < len; i++) {            num[i] = scanner.nextInt();        }        System.out.println("the longestConsecutive num is:"                + longestConsecutive2(num));    }    public static int longestConsecutive1(int[] num) {        int len = num.length;        if (len == 0) {            return 0;        }        Arrays.sort(num);        int count = 1;        int maxLen = 1;        for (int i = 1; i < len; i++) {            if (num[i] == num[i - 1] + 1) {                count++;            } else {                if (num[i] != num[i - 1]) {                    count = 1;                }            }            if (count > maxLen) {                maxLen = count;            }        }        return maxLen;    }    public static int longestConsecutive2(int[] num) {        int len = num.length;        if (len == 0) {            return 0;        }        Set<Integer> set = new HashSet<Integer>();        int maxLen = 1;        for (int i = 0; i < len; i++) {            set.add(num[i]);        }        for (int i = 0; i < len; i++) {            int count = 1;            int leftNum = num[i] - 1;            int rightNum = num[i] + 1;            while (set.contains(leftNum)) {                count++;                set.remove(leftNum);                leftNum--;            }            while (set.contains(rightNum)) {                count++;                set.remove(rightNum);                rightNum++;            }            if(count > maxLen){                maxLen = count;            }        }        return maxLen;    }}