leetcode-single number
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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:
a^a=0; a^0=a;将所有元素进行异或
代码:
int singleNumber(int A[], int n) {
int res=0;
for(int i=0; i<n; ++i)
{
res=res^A[i];
}
return res;
}
0 0
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