1 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, 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].2 尝试解
2.1 分析
给定一串无重复的整数数组，有且仅有一对数的和为target，问这两个数在数组中的序列值。
直觉想法是先将数组排序，然后用双指针法查找这两个数，然后再查找这两个数的序列值。时间复杂度为O(n log n)。
2.2 代码
class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> copy = nums; sort(nums.begin(),nums.end()); int left = 0, right = nums.size()-1; while(left < right){ if(nums[left]+nums[right]<target) left++; else if(nums[left]+nums[right]>target) right--; else break; } vector<int> result; for(int i = 0; i < copy.size();i++){ if(copy[i]==nums[left] || copy[i]==nums[right]) result.push_back(i); } return result; } };3 标准解
3.1 分析
由于所有数无重复，且target已知，可以用哈希表将每个数的值及其序列存储起来，然后对于每一个数num，查找target-num是否在哈希表中。
3.2 代码
class Solution { public: 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]); result.push_back(i); return result; } //number was not found. Put it in the map. hash[numbers[i]] = i; } return result; } };
