题解 | #两数之和#

python 暴力法：

        for i in range(len(numbers)):
            for j in range(i+1, len(numbers)):
                if numbers[i] + numbers[j] == target:
                    return i+1, j+1

python 预先准备好map，由于数字可能重复出现所以map是{6:[1,4], 7:[2]}的形式，数组内表示数字出现的次数

        map = {}
        for i, n in enumerate(numbers):
            if map.get(n):
                map[n].append(i+1)
            else:
                map[n] = [i+1]

        for n in numbers:
            if target - n in map:
                if map[n][0] != map[target-n][0]:
                    return map[n][0], map[target-n][0]
                elif len(map[target-n]) > 1:
                    return map[n][0], map[target-n][1]

python 在遍历中创建map（三个中的最优）

        map = {}
        for i, n in enumerate(numbers):
            if map.get(target - n) is not None:
                return map[target-n]+1 , i+1
            else:
                map[n] = i