题解 | #数组中重复的数字#

空间复杂度O(n/32)，时间复杂度O(n)

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param numbers int整型一维数组 
     * @return int整型
     */
    public int duplicate (int[] numbers) {
      // write code here
        if (numbers == null || numbers.length < 2) {
            return -1;
        }
        int maxN = numbers.length;
        int[] bitMap = new int[maxN / 32 + 1];
        for (int number : numbers) {
            int position = number / 32;
            int bitPos = number & 31;
            int bit = bitMap[position] >> bitPos;
            if ((bit & 1) == 1) {
                return number;
            } else {
                bitMap[position] = bitMap[position] | (1 << bitPos);
            }
        }
        return -1;

        
    }
}