feat: update solution to lcci problem: No.03.02 (#2323)

Author: yanglbme

lcci/03.02.Min Stack/README_EN.md

@@ -28,7 +28,16 @@ minStack.getMin();   --&gt; return -2.</pre>
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Double Stack
+
+We use two stacks to implement this, where `stk1` is used to store data, and `stk2` is used to store the current minimum value in the stack. Initially, `stk2` stores a very large value.
+
+-   When we push an element `x` into the stack, we push `x` into `stk1`, and push `min(x, stk2[-1])` into `stk2`.
+-   When we pop an element from the stack, we pop the top elements of both `stk1` and `stk2`.
+-   When we want to get the top element in the current stack, we just need to return the top element of `stk1`.
+-   When we want to get the minimum value in the current stack, we just need to return the top element of `stk2`.
+
+For each operation, the time complexity is $O(1)$, and the space complexity is $O(n)$.
 
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