LeetCode - Refresh - Single Number
2015-03-23 15:05
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You can use a hash map to record the frequencys. Or you can use bit operation.
x ^ x = 0. So the only left is the one.
x ^ x = 0. So the only left is the one.
class Solution { public: int singleNumber(int A[], int n) { int result = 0; for (int i = 0; i < n; i++) { x ^= A[i]; } return result; } };
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