leetcode 1 Two Sum-优快云博客

本文介绍了一种解决两数之和问题的有效算法。通过排序和双指针技术找到两个数，使其和为目标值，并返回这两个数在原始数组中的位置。此算法复杂度为O(nlogn)，适用于寻找唯一解的情况。

Given an array of integers,find two numbers such that they add up to a specific target number.

The function twoSum shouldreturn indices of the two numbers such that they add up to the target, whereindex1 must be less than index2. Please note that your returned answers (bothindex1 and index2) are not zero-based.

You may assume that eachinput would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

思路：先拷贝numbers，然后再将该副本排序，然后左右夹逼，排序O(n log n)，左右夹逼O(n)，最终O(n log n)。再把可能的数字在原数组中查找其index

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        vector <int> temp;
		vector <int> ans;
		vector<int>::iterator it1;
		vector<int>::iterator it2;
		int i;
		int num;

		temp.resize(numbers.size());//这一步必不可少

		copy(numbers.begin(),numbers.end(),temp.begin());
		sort(temp.begin(),temp.end());

		it1=temp.begin();
		it2=temp.end()-1;//特别注意需要减一

		while(it1!=it2)
		{
			if (( *it1 + *it2 )== target)
			{  
			    num=*it1;
			    break;
			}
			else if(*it1+*it2>target) it2--;
			else it1++;
		}
		for(i=0;i<numbers.size();i++)
		{
			if(numbers.at(i)==num||numbers.at(i)==(target-num)) 
			{
				ans.push_back(i+1);
				cout<< i<<endl;
			}
		}
		return ans;
    }
};