[Leetcode 136, Medium] Single Number I
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Problem:
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?
Analysis:
算法:逻辑运算的异或(^)是交换且可结合的,而且,是幂等的。所以,如果是有且只有一个单独出现的元素,那么把所有的元素进行异或操作,所得到的值就是所要的结果。
衍生问题:every element appears even times except for one which appears odd times.
Solution:
C++:
(I)
int singleNumber(int A[], int n) { if(n == 0) return 0; int rval = A[0]; for(int i = 1; i < n; ++i) rval ^= A[i]; return rval; }
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