go] [array] 1. Two Sum
原題連結
解法一：with map
用map來儲存已經出現過的數字，及它對應的下標，每遍歷到一個新的數字，就判斷 target - nums[i] 是否已經存在，若存在，則返回它們對應的下標，否則，將當前數字及其下標存入map, 繼續遍歷。
時間複雜度 : O(N), 空間複雜度: O(N)
這是最直接，也最容易想到的解法。當我用go去實現它的時候，執行時間還是嚇了我一跳
解法二： brute force 暴力解法，對於陣列中的每一個數字，遍歷其他數字，判斷它們只和是否為 target.
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