Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
题目链接：https://leetcode.com/problems/two-sum/description/
解法一：两重循环遍历数组（时间复杂度：O(n2)）
解法二：1.用map来存放，键值为数组的值，实值为个数，
2.遍历数组，用find查找是否存在我们要找的值（target - nums[i]）,找到判断是否与nums[i]相等，相等则判断是否有两个同样的值
3.最后记录下标值的时候，要注意当找到的两个值相等的时候，不能记录为相同的，要记录下一个
时间复杂度O(nlog(n))，其中O(log(n))为map容器产生
vector<int> twoSum(vector<int>& nums, int target) { int length = nums.size(); vector<int> vec; map<int , int> simap; int num = 0; for(int i = 0; i < length; i++) simap[nums[i]] ++; for(int i = 0; i < length; i++) { num = target - nums[i]; if(simap.find(num) != simap.end())//找到了 { if(nums[i] == num && simap[num] == 1)//同一个值 continue; vec.push_back(i); break; } } if(vec.size() != 0)//找到了和为target的两个数 { for(int i = 0; i < length; i++) if(nums[i] == num) { if(vec[0] == i) continue; vec.push_back(i); break; } } return vec; }
解法三：1.vec存储pair类型，分别为数组的值和下标，
2.存储完后用sort进行排序，
3.用迭代器遍历数组，调用lower_bound函数，查找另一个值(target - ite->first)
时间复杂度(O(nlog(n)))
vector<int> twoSum(vector<int>& nums, int target) { int length = nums.size(); vector<pair<int , int> > vec; vector<int> res; for(int i = 0; i < length; i++) { vec.push_back(make_pair(nums[i],i)); } sort(vec.begin(),vec.end()); vector<pair<int , int>>::iterator tmp,i,ite = vec.begin(); while(ite != vec.end()) { auto tm = next(ite, 1); i = lower_bound(tm,vec.end(),make_pair(target - (ite->first), 0), [](const pair<int, int> & a, const pair<int, int> & b) -> bool { return a.first < b.first; }); if(i!=vec.end()&&i->first + ite->first == target) { res.push_back(ite->second); res.push_back(i->second); break; } ite ++; } return res; }
