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LeetCode-Two Sum
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
思路：很简单的想法就是用两层循环，遍历所有进行匹配。但是这种方法是非常不高效的。
高效的方法（nlogn）:利用map数据结构，逐个遍历，计算target与当前值numbers[i]的差，如果在map中存在该值，则说明匹配。如果不存在该值，将numbers[i]，i 插进map数据结构中。map存储结构是<numbers[i], i>
1 class Solution { 2 public: 3 vector<int> twoSum(vector<int> &numbers, int target) { 4 int i, sum; 5 vector<int> ret; 6 map<int, int> hashmap; 7 for(i=0; i<numbers.size(); i++) { 8 if(!hashmap.count(numbers[i])) { 9 hashmap.insert(pair<int, int>(numbers[i], i));10 }11 12 if(hashmap.count(target-numbers[i])) {13 int n = hashmap[target-numbers[i]]; //索引值14 if(n < i){15 ret.push_back(n+1);16 ret.push_back(i+1);17 return ret;18 }19 }20 }21 22 return ret;23 }24 };
LeetCode-Two Sum
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