【Leetcode-easy-136】Single Number
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【Leetcode-easy-136】Single Number
题目
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?
思路
位运算——异或运算
只有一个1则结果为1,也即是“有1则1”。
则可以推出:
- 任何整数和0异或结果是它本身。
- 一个整数异或它本身结果等于0。
可以进一步推出:
- 一个整数异或另一个整数两次结果是它本身。
程序
class Solution { public int singleNumber(int[] nums) { int result = 0; for (int num : nums) result ^= num; return result; }}
声明:题目来自Leetcode。
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