Two Sum (leetcode)
2014-10-10 17:33
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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
https://oj.leetcode.com/problems/two-sum/ 解法:
使用hash map 保存vector中的数值和下标, <key, value> 对应 <数值,下标>, 然后遍历整个vector, 查找是存在满足条件的数值
时间复杂度: O(n)
空间复杂度: O(n)
code:
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
https://oj.leetcode.com/problems/two-sum/ 解法:
使用hash map 保存vector中的数值和下标, <key, value> 对应 <数值,下标>, 然后遍历整个vector, 查找是存在满足条件的数值
时间复杂度: O(n)
空间复杂度: O(n)
code:
typedef std::unordered_map<int, int> map_type; typedef std::unordered_map<int, int>::iterator map_iterator; class Solution { public: vector<int> twoSum(vector<int> &numbers, int target) { map_type map; std::vector<int> ret_vec; int size = numbers.size(); //using hashmap to keep the connection between value and it's address index for (int i = 0; i != size; ++i) { map[numbers[i]] = i; } for (int i = 0; i != size; ++i) { map_iterator it = map.find(target - numbers[i]); if (it != map.end()) { if (i == it->second) { continue; } else if (i < it->second) { ret_vec.push_back(i + 1); ret_vec.push_back(it->second + 1); } else { ret_vec.push_back(it->second + 1); ret_vec.push_back(i + 1); } return ret_vec; } } return ret_vec; } };
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