LeetCode136 Single Number
- Given an array of integers, every element appears twice except for one. Find that single one.
- Notice:Your algorithm should have a linear runtime complexity. Implement it without using extra memory.
- Analysis:
- The problem itself is not hard. However, the additional time and spcace overhead limitation is a bit annoying. I have no clue at the first sight of this probelm,either. I finally solve it by referring to others' solutions, which reminds me that xor operation is the key to the problem.
- We all know A^A equals to zero, then we can xor all the numbers and the final answer will be the number appeaing just once.
- Code:
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1 class Solution { 2 public: 3 int singleNumber(int A[], int n) { 4 int ans = 0; 5 for (int i = 0; i < n; i++){ 6 ans ^= A[i]; //xor 7 } 8 return ans; 9 } 10 };
- Actually, the code works well not only for appearing twice, but also appearing even times.
- In next bolg, I will share you will a more general way to handle this kind of question.
posted on 2015-04-10 17:09 Jasonzhang398 阅读(329) 评论(0) 编辑 收藏 举报