05 August 2008
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:
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Input: [2,2,1]
Output: 1
Example 2:
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Input: [4,1,2,1,2]
Output: 4
给一个数组，每个数字出现两次，找出只出现过一次的数字。中规中矩用HashMap来统计次数自然是没什么难度，这里用位操作中XOR，当两个数相同的时候为0，位数相互进行XOR后，最后的结果一定会是只出现一次的元素。
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class Solution {
public int singleNumber(int[] nums) {
int result = 0;
for (int num : nums) {
result ^= num;
}
return result;
}
}