寻找一个数字,重复甚至没有的时候,所有的其他号码重复的次数多无 [英] Finding a number that repeats even no of times where all the other numbers repeat odd no of times
问题描述
由于是整数数组。阵列中的每个编号重复的次数为奇数,但是只有1号被重复的偶数次。找到这个数字。
Given is an array of integers. Each number in the array repeats an ODD number of times, but only 1 number is repeated for an EVEN number of times. Find that number.
我在想一个哈希表,每个元素的数量。它需要O(n)的空间。有没有更好的办法?
I was thinking a hash map, with each element's count. It requires O(n) space. Is there a better way?
推荐答案
哈希地图是好的,但你需要存储的每个元素的数模2。 所有这些都将最终被1(奇)除0(偶数)-count元素。
Hash-map is fine, but all you need to store is each element's count modulo 2. All of those will end up being 1 (odd) except for the 0 (even) -count element.
(如阿莱克斯摹说,你并不需要使用运算(数+%2
),只有XOR(计数^ =为0x1
);但任何编译器将优化反正)
(As Aleks G says you don't need to use arithmetic (count++ %2
), only xor (count ^= 0x1
); although any compiler will optimize that anyway.)
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