LeetCode | 1) Two sum
2016-10-02 15:51
260 查看
题目
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.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { }; }
思路
方案 1:两次循环遍历整个数组,检测nums[i] + nums[j] == target,时间复杂度O(N2)
方案 2:首先利用
map<int, size_t>将数据和对应的下标存储起来;然后一次遍历数组,在hash map中查找
tartget - nums[i],时间复杂度是O(n)∗O(1)=O(N). 该方法需要两次hash,一次是建立hash表,另一次是查找hash表,存在有可能找到的数和num[i]一样这个问题
方案 3:只用一遍hash表。首先查找hash表中是否存在
nums[i],不存在则将
target - nums[i]存储在hash表中。
代码
class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> vec; unordered_multimap<int, size_t> numMap; //元素之间没有序关系,宜用unordered_map; nums数组中有多个同样的数,宜用multimap //综合考虑,使用unordered_multimap for (size_t it = 0; it != nums.size(); ++it) { auto iter = numMap.find(nums[it]);//先对numMap进行查找 if (iter == numMap.end()) { numMap.insert({target - nums[it], it});//没有找到则插入数据和下标 } else { return vector<int>{iter->second, it}; //找到,直接返回numMap的迭代器iter指向的下标(iter->second)和当前nums的下标it。 } } return vec; } };
运行时间:19ms
相关文章推荐
- LeetCode笔记-A1-Two Sum
- 【LeetCode】1 - Two Sum
- leetcode1~Two Sum
- LeetCode 1 Two Sum 解题报告
- LeetCode 1 Two Sum
- [leetcode 1] Two Sum
- Leetcode 1. Two Sum
- Leetcode[1] Two Sum
- [leetcode] 1.Two Sum
- [LeetCode] Two Sum
- [leetcode] 1. Two Sum
- LeetCode - Two Sum
- LeetCode 1.Two Sum
- LeetCode Two Sum 及C++map浅显理解
- leetcode---------------Two Sum
- leetcode 1 Two Sum
- [leetcode, python] Two Sum 两数之和等于某数
- LeetCode——two sum
- Leetcode 1. Two Sum
- LeetCode1-Two Sum