Leetcode Single Number

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?

对于这道题，要求首先是线性的，其次最好不用到额外的空间，那么我们就用一个set来存储数字，同时用一个变量来记录。首先遍历set，若数组中这个数字已经出现了，则将变量num减去这个数字，若set中没有该数字，则将变量set+该数字，最终结果变量Num的值会等于那个出现一次的数字。

public class Solution {
    public int singleNumber(int[] A) {
  int single = 0;
    Set<Integer> s = new TreeSet<Integer>();
    for (int i : A) {
      if (s.contains(i)) {
        single = single - i;
        s.remove(i);
      } else {
        s.add(i);
        single = single + i;
      }
    }
  return single;
    }
}

Leetcode Single Number

原文：http://www.cnblogs.com/criseRabbit/p/4120747.html