1. The time complexity of the brute force solution is`O(n**2)`. To improve efficiency, you can sort the array, and then use**two pointers**, one pointing to the head of the array and the other pointing to the tail of the array, and decide`left += 1` or`right -= 1` according to the comparison of`sum` and`target`.

2. After finding the two values which`sum` is`target`, you can use the`index()` method to find the`index` corresponding to the value.

2. After sorting an array of numbers, if you want to know the original`index` corresponding to a certain value, there are two solutions:

- Solution 1: Use`index()` method to find it.

- Solution 2: Bring the`index` when sorting, that is, the object to be sorted is an array of tuples of`(num, index)`.

* Time:`O(N * log N)`.

1. 暴力解法的时间复杂度为`O(n**2)`，想提升效率，可以对数组进行排序，然后用双指针，一个指向数组头，一个指向数组尾，根据**和**情况决定`left += 1`还是`right -= 1`。

2. 找出了两个值后，需要用`index()`方法去找值对应的`index`。

2. 对数值数组排序后，想知道某个数值对应的原来的索引下标，有两种方案：

- 方案1：使用index() 查找；

- 方案2：在排序时带上索引下标，即排序的对象是元组`(num, index)`的数组。

1.`Map`中，`key`是`num`，`value`是数组`index`。