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?
class Solution {
public:
int singleNumber(int A[], int n) {
int sigleNum = 0;
for (int i = 0; i < n; i++)
{
sigleNum^= A[i];
}
return sigleNum;
}
};
本文介绍了一种算法,该算法能在给定的整数数组中找出仅出现一次的整数,其余元素都出现了两次。算法实现了线性运行时间复杂度,并且不使用额外内存。文中提供了一个名为Solution的类及其实现方法singleNumber。

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