leetcode解题: Single Number (136)

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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?

解法1：O(N) time with O(1) space, N is the number of elements

运用XOR是一个抵消运算符，A XOR A 出来是一个0，所以对所有的数字做XOR之后重复的都消掉了，只剩下单独的一个。C++中vector的method是vector.size(),和java中array的array.length注意区分。

C++

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class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int result = nums[0];
        for (int i = 1; i < nums.size(); i++) {
            result ^= nums[i];
        }
        return result;
    }
};

Java

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public class Solution {
    public int singleNumber(int[] nums) {
        int result = nums[0];
        for (int i = 1; i < nums.length; i++) {
            result ^= nums[i];
        }
        return result;
    }
}