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?
解题思路:
按位异或运算,出现两次的都归零了,剩下的就是出现一次的了,JAVA实现如下:
public int singleNumber(int[] nums) { for(int i=1;i<nums.length;i++) nums[0]^=nums[i]; return nums[0]; }
时间: 2024-11-09 20:29:02