leetcode Question: Single Number I

Single Number

Given an array of integers, every element appears twice except for one. Find that single one.
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?


The requirement is O(n) time and O(1) space.
Thus, the  "first sort and then find " way is not working.
Also the "hash map" way is not working.

Since we can not sort the array, we shall find a cumulative way, which is not about the ordering.

XOR is a good way, we can use the property that A XOR A = 0, and A XOR B XOR A = B.

So, the code becomes extremely easy.


class Solution {
    int singleNumber(int A[], int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int res = A[0];
        for (int i=1;i<n;i++){
            res = res ^ A[i];
        return res;

Code (Python):

class Solution:
    # @param A, a list of integer
    # @return an integer
    def singleNumber(self, A):
        res = 0
        for a in A:
            res = res ^ a
        return res

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