[leetcode] 260. Single Number III
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260. Single Number III
Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.
For example:
Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].
Note:
- The order of the result is not important. So in the above example,
[5, 3] is also correct. - Your algorithm should run in linear runtime complexity. Could you
implement it using only constant space complexity?
题意:
给定一个数字数组nums,其中两个元素只出现一次,并且所有其他元素出现两次。找到两个只出现一次的元素。
方法一
思路:
利用map进行遍历统计元素个数,map的键存储元素本身,值存储元素出现的次数。遍历map,如果值不为2,则此元素的键就是所查找的元素的值。代码如下:
vector<int> singleNumber(vector<int>& nums) { map<int,int> m; vector<int> num; int n=nums.size(); for (int i=0;i<n;i++) { if(m.find(nums[i])==m.end()) m[nums[i]]=1; else m[nums[i]]++; } for (auto itr=m.begin();itr!=m.end();++itr) { if (itr->second!=2) { num.push_back(itr->first); } } return num; }
方法二
思路:
先排序,然后判断相邻的三个元素是否相同,不同的即为单元素,但是要注意当比较数组两端的元素的时候会出现数组越界的问题,所以要先单独比较两端的元素。先判断不同元素是否在vector的两端,如果在返回,如果不在再一次比较数组除去两头元素后,中间元素相邻的三者间是否相同,不同则返回。代码如下:
vector<int> singleNumber(vector<int>& nums) { vector<int> num; sort(nums.begin(),nums.end()); int n=nums.size(); if (nums[0]!=nums[1]) { num.push_back(nums[0]); } if(nums[n-1]!=nums[n-2]){ cout<<"jinqy"<<endl; num.push_back(nums[n-1]); } for (int i=1;i<n-1;i++) { if (nums[i]!=nums[i-1]&&nums[i]!=nums[i+1]) { cout<<"jinqy2"<<endl; num.push_back(nums[i]); } } return num; }
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