剑指offer - 数组中重复的数字

Method 1 Time:O(n) Space:O(1)
Method 2 Time:O(n) Space:O(n)
Special Test Case: [6,3,2,0,2,5,0]

public class Solution {
    public boolean duplicate(int numbers[],int length,int [] duplication) {
        if(numbers == null || length <= 0) {
           return false;
        }
        for(int i = 0; i < length;i++) {
            int index = numbers[i];
            if(index < 0 || index >= length) {
                return false;
            }
            if(index == i) {
                continue;
            }
            if(numbers[index] == index) {
                duplication[0] = index;
                return true;
            }else{
                numbers[i] = numbers[index];
                numbers[index] = index; // swap once
            }
        }
        return false;
    }
}

import java.util.*;
public class Solution {
    public boolean duplicate(int numbers[],int length,int [] duplication) {
        if(numbers == null || length <= 0) {
           return false;
        }
        LinkedHashSet<Integer> set = new LinkedHashSet<>();
        for(int i = 0;i < length;i++) {
            int val = numbers[i];
            if(val < 0 || val >= length) {
                break;
            }else if(set.contains(val)) {
                duplication[0] = val;
                return true;
            }
            set.add(val);
        }
        return false;
    }
}