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, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
vector<int> twoSum(vector<int>& nums, int target) { if (nums.size() == 0) return nums; vector<int>result; //sort(nums.begin(), nums.end()); for (int i = 0; i < nums.size(); i++) { for (int j = i + 1; j < nums.size();j++) { if (nums[i]+nums[j] == target) { result.push_back(i); result.push_back(j); return result; } } } return result; }
然后看到大牛的代码：只需遍历一次数组，时间复杂度为O(n).
vector<int> twoSum(vector<int> &numbers, int target) { //Key is the number and value is its index in the vector. unordered_map<int, int> hash; vector<int> result; for (int i = 0; i < numbers.size(); i++) { int numberToFind = target - numbers[i]; //if numberToFind is found in map, return them if (hash.find(numberToFind) != hash.end()) { //+1 because indices are NOT zero based result.push_back(hash[numberToFind] + 1); result.push_back(i + 1); return result; } //number was not found. Put it in the map. hash[numbers[i]] = i; } return result; }
原文：http://www.cnblogs.com/hellowooorld/p/6431329.html