137 Single Number II

weixin_39145266

于 2019-04-12 17:58:27 发布

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分类专栏： # BitManipulation # Solved LeetCode # Medium

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本文探讨了一种在整数数组中找出唯一出现一次元素的算法，该算法具备线性时间复杂度且几乎不使用额外内存。通过对比尝试解与标准解，深入解析了位操作技巧在解决此类问题上的高效应用。

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1 题目

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

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

Example 1:

Input: [2,2,3,2]
Output: 3
Example 2:

Input: [0,1,0,1,0,1,99]
Output: 99

2 尝试解

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        for(int i = 0; i < nums.size();i++){
            if((i > 0 && nums[i-1] == nums[i])||(i < nums.size()-1 && nums[i]==nums[i+1]))
               continue;
            else
               return nums[i];
        }
        return 0;
    }
};

3 标准解

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int x1 = 0, x2 = 0, mask = 0;
         
        for (int i : nums) {
            x2 ^= x1 & i;
            x1 ^= i;
            mask = ~(x1 & x2);
            x2 &= mask;
            x1 &= mask;
        }

        return x1;
    }
};