兩數之和
給定一個整數數組 nums 和一個目標值 target，請你在該數組中找出和為目標值的那 兩個 整數，并返回他們的數組下標。

你可以假設每種輸入只會對應一個答案。但是，數組中同一個元素不能使用兩遍。

示例:

給定 nums = [2, 7, 11, 15], target = 9

因為 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]

解題思路：
· for j := i+1;j < k;j++ ·j肯定大于i所以不需要在判斷j的有效性，因此效率更高

func twoSum(nums []int, target int) []int {k := len(nums)for i := 0; i < k; i++ {tmp := nums[i]for j := i+1;j < k;j++ {if target == tmp + nums[j] {return []int{i, j}}}}return nil } vector<int> twoSum1(vector<int>& nums, int target) {int n = nums.size();for (int i = 0; i < n; i++){for (int j = i+1;j < n; j ++){if (nums[i] + nums[j] == target)return {i, j};}}return {}; }

使用C++求解：
解題思路，新建一個hash表， map鍵值對，將nums[i]作為hash表的key，對應的索引作為value，我們所要作的就是，在hash表中查找對應的target - nums[i]的key是否存在，若是存在就返回當前的value和對應的i，要是不存在就接著將對應nums[i]和索引按照鍵值對存入map，代碼實現如下：

vector<int> twoSum2(vector<int>& nums, int target) {unordered_map<int, int> hashTable;for (int i = 0;i < nums.size(); i++) {auto it = hashTable.find(target - nums[i]);if(it != hashTable.end())return {it->second, i};hashTable[nums[i]] = i;}return {}; } func twoSum(nums []int, target int)[]int {hashTable := map[int]int{}for i, x := range nums {if p, ok := hashTable[target - x]; ok {return []int{p, i}}hashTable[x] = i}return []int{} } // // Created by andrew on 2021/3/5. // #include <iostream> #include <vector> #include <map> #include <unordered_map>using namespace std;// 暴力求解 class Solution { public:vector<int> twoSum1(vector<int>& nums, int target) {int n = nums.size();for (int i = 0; i < n; i++){for (int j = i+1;j < n; j ++){if (nums[i] + nums[j] == target)return {i, j};}}return {};}vector<int> twoSum2(vector<int>& nums, int target) {unordered_map<int, int> hashTable;for (int i = 0;i < nums.size(); i++) {auto it = hashTable.find(target - nums[i]);if(it != hashTable.end())return {it->second, i};hashTable[nums[i]] = i;}return {};}};int main(int argc, char *argv[]) {vector<int> nums = {2,7,11,15};vector<int> retVal = {};int target = 9;Solution solution;retVal = solution.twoSum1(nums, target);int n = retVal.size();for (int i = 0; i < n; i ++) {cout << retVal[i] << endl;}cout << "+++++++++++++++++++++++++++++++" << endl;retVal = solution.twoSum2(nums, target);n = retVal.size();for (int i = 0; i < n; i ++) {cout << retVal[i] << endl;}return 0; } package mainimport "fmt"func twoSum(nums []int, target int)[]int {hashTable := map[int]int{}for i, x := range nums {if p, ok := hashTable[target - x]; ok {return []int{p, i}}hashTable[x] = i}return []int{} }func main() {var nums = []int{2, 3, 7, 11, 15}var target int = 9retVal := twoSum(nums, target)fmt.Println(retVal) } 與50位技術專家面對面20年技術見證，附贈技術全景圖

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