SingleNumber [source code]

public class SingleNumber {
    public int singleNumber(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i : nums) {
            Integer result = map.get(i);
            if (result == null) map.put(i, 1);
            else map.put(i, -1);
        }
        for (Map.Entry e : map.entrySet())
            if ((int) e.getValue() == 1) return (int) e.getKey();
        return 999;
    }

    public static void main (String[] args) {
        SingleNumber tester = new SingleNumber();
        int[] input1 = {2,2,1,3,3};
        assert tester.singleNumber(input1) == 2 : "fail 1";
        int[] input2 = {1,2,3,5,1,3,2,6,6};
        System.out.println(tester.singleNumber(input2));
    }
}

这个算法还是比较简单直接的, 就是最后的速度不怎么样, 21ms, 10%. 看看答案有没有什么更好的方法;

Problem Description

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Hide Tags Hash Table Bit Manipulation
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