136. 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?
采用异或运算可以不需要额外的内存,并且可以过一遍线性时间内运算完:
class Solution {public: int singleNumber(vector<int>& A) { int size = A.size(); for (int i = 1; i < size; ++i) A[0] ^= A[i]; return A[0]; }};
也可以估算其他方案:
1、最基本的暴力遍历。开辟一个变量记录当前值,与之后的对比,最坏情况O(n2)
2、开辟一个数组,遍历到一个就把对应位置加一,该方法问题空间分配大小不确定,假设确定的话,那么线性时间O(n)可以解决
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