leetcode-python-java
diff --git a/‎en/1-1000/1-two-sum.md‎
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diff --git a/‎zh/1-1000/1-two-sum.md‎
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@@ -44,7 +44,11 @@ The second train of thought is, without changing the array, can we use additiona
 
 ## Intuition
 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.
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+2. After sorting an array of numbers, if you want to know the original `index` corresponding to a certain value, there are two solutions:
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+    - 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)`.
 
 ### Complexity
 * Time: `O(N * log N)`.
 
@@ -49,7 +49,11 @@ The second train of thought is, without changing the array, can we use additiona
 ## 思路
 ### 思路1：双指针
 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`。
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+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。
 
 ### 思路2：使用 Map 提升查找一个值的效率
 1. `Map`中，`key`是`num`，`value`是数组`index`。