LeetCode专题：136.Single Number

Single Number

Given a non-empty 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?
Example 1:
Input: [2,2,1]
Output: 1
Example 2:
Input: [4,1,2,1,2]
Output: 4

解答

/**
 Given a non-empty 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?
 Example 1:
 Input: [2,2,1]
 Output: 1
 Example 2:
 Input: [4,1,2,1,2]
 Output: 4
 */
public class SingleNumber {
    public static int singleNumber(int[] nums) {

        /**方法一：最直观的想法，把数组中的所有数遍历一遍，从第一个开始，看有几个相同。
         * 这样写复杂度为o(n^2)复杂度太高。 有兴趣的小伙伴可以自己动手写一遍
         */

        /**
         * 方法二借助hashmap遍历一边数组，记录数组中出现的次数，最后取出value为一的key
         * 这里也省了，小伙伴自己动手写一遍
         */

        /**
         * 这里主要讲解方法三：使用异或运算符
         * 举个例子：如果 a^b^a = b
         *   a^b^a^b^c = c
         *   1^1^2=2
         *   这样一来就很简单了
         */
        int p=nums[0];
        for (int i=1;i<nums.length;i++){
            p^=nums[i];
        }
        return p;
    }

    public static void main(String[] args) {
        int [] nums={5,6,6,3,8,3,8,5,7};
        System.out.println(singleNumber(nums));
    }
}