题目:
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
解题思路:
使用HashMap存储数组元素值与下标的映射,遍历数组,如果符合要求的元素已经存在于HashMap中,则返回结果;否则将当前元素加入HashMap。算法时间复杂度O(n),空间复杂度O(n)。
class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, int> mapping; vector<int> result; for (int i=0;i<nums.size();i++) { if (mapping.find(target-nums[i]) == mapping.end()) { mapping[nums[i]] = i; } else { int former = mapping[target-nums[i]]; result.push_back(former+1); result.push_back(i+1); break; } } return result; } };
原文:http://www.cnblogs.com/wwwjjjfff/p/4851678.html