LeetCode刷题之第一题——TwoSum

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


The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.


You may assume that each input would have exactly one solution.


Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

注意1.索引号从1开始

       2.只有一个解

解题思路

A方案（两层循环）O(n2)

class Solution {public:    vector<int> twoSum(vector<int>& numbers, int target) {        vector<int> result;         for (int i = 0; i < numbers.size()-1; i++) {             for (int j = i+1; j < numbers.size(); j++) {                 if (numbers[i] + numbers[j]==target) {                     result.push_back(i);                     result.push_back(j);                     return result;                 }             }         }         return result;            }};

B方案O(nlogn)。排序，然后两个指针一前一后。因为题中说明了只有一对答案，因此不需要考虑重复的情况。

c方案 O(nlogn)用map结构来完成

map的相关介绍：http://blog.sina.com.cn/s/blog_61533c9b0100fa7w.html

class Solution {public:   vector<int> twoSum(vector<int> &numbers, int target) {        vector<int> result;        map<int, int> m;        if (numbers.size() < 2)            return result;        for (int i = 0; i < numbers.size(); i++)            m[numbers[i]] = i;        map<int, int>::iterator it;        for (int i = 0; i < numbers.size(); i++) {            if ((it = m.find(target - numbers[i])) != m.end())            {                if (i == it->second) continue;                result.push_back(i);                result.push_back(it->second);                return result;            }        }         return result;            }};