Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
[Thoughts]
不知道这个题从哪里来的,但是明显是针对计算机专业的。很简单,就是位操作,任意两个相同的数如果做异或(Exclusive Or)运算的话,结果为0.所以,这题的解法就是这么直白,从0开始到n,一路异或下去,最后剩下的值就是所求。
[Codes]
1: int singleNumber(int A[], int n) {
2: int left = A[0];
3: for(int i =1; i< n; i++)
4: {
5: left = left ^ A[i];
6: }
7: return left;
8: }
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