LeetCode:Single Number
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Single Number
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?
<span style="font-size:18px;">int singleNumber(int A[], int n){ int result = 0; for(int i = 0;i<n;i++) result ^= A[i]; return result;}</span>偶数个相同的整数异或后为0，所以遍历数组，与所有的数组元素相继异或，最后异或的结果就是单独 的整数已AC 36ms
0 0
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