leetcode Bit Manipulation

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最新推荐文章于 2025-08-09 17:01:50 发布

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分类专栏： algorithm Leetcode 文章标签： algorithm leetcode C

本文链接：https://blog.youkuaiyun.com/hugoR10K/article/details/43339817

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本文介绍了一种使用位操作解决问题的方法，具体为在一个整数数组中找到仅出现一次的整数，而其他元素均出现两次或三次。算法具备线性时间复杂度且不使用额外内存，适用于SingleNumber I和II两种情况。

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Bit Manipulation, work from latched

Single Number I

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?

Note ： not use python build in set function, which can return unique number in the list

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

Single Number II

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

class Solution {
public:
    int singleNumber(int A[], int n) {
        int oneNum = 0;
        int twoNum = 0;
        int threeNum = 0;
        for(int i = 0 ; i < n ;i++){
            threeNum = twoNum & A[i];
            twoNum = oneNum & A[i] | twoNum;
            oneNum = oneNum | A[i];
            oneNum = oneNum &(~threeNum);
            twoNum = twoNum & (~threeNum);
            threeNum = 0;
        }
        return oneNum;
    }
};

Summary : or operation to get 1, and operation to accumulate 1, and (not) to clear.

This method can be extend for more case for question like single number