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Given an array of integers, every element appears twice except for one. Find that single one.
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Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
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int singleNumber(vector<int>& nums) { unordered_map<int, int> cnt; for (auto i : nums) { if (cnt.find(i) != cnt.end()) cnt.erase(i); else cnt.insert(make_pair(i, 1)); } return (*cnt.begin()).first; }
去discuss中，发现还有更简单的解法，使用XOR位运算，因为A XOR A = 0，而且XOR具有结合性，所以A XOR B XOR C XOR A XOR C XOR B = 0
int singleNumber(vector<int>& nums) { int result = 0; for (int i = 0; i < nums.size(); i++) { result ^= nums[i]; } return result; }
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原文地址：http://www.cnblogs.com/sdlwlxf/p/5143278.html
