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?
public int singleNumber(int[] nums) {int result = 0;for (int i = 0; i < nums.length; i++) {result ^= nums[i];}return result;}
注意到A^A=0,所以A^B^A=B
0 0
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