C++实现O(1)时间找栈最小元素：双最小值变量法的失效场景与问题

Hey there! Let's dig into why tracking just min1 and min2 to get O(1) minimum lookups in a stack doesn't hold up in all scenarios, and what specific flaws exist in this approach.

These are the specific cases where your implementation will break and return incorrect minimum values:

• Multiple instances of the minimum value
  Suppose you push elements in this order: [3, 1, 1, 2]. Your min1 would be 1, min2 would also be 1 (since the second 1 doesn't change the top two minima). Now, if you pop both 1s, the stack is left with [3, 2], where the actual minimum is 2. But your logic would set min1 to min2 (still 1) after each pop, so min() would incorrectly return 1 instead of 2.

• Min2 gets popped before min1
  Let's say you push elements: [5, 4, 3, 2, 1]. Here, min1 = 1, min2 = 2. Now pop 2 (which is min2, not min1). Your code doesn't update min2 in this case, so min2 stays 2 even though it's no longer in the stack. Next, when you pop 1 (min1), your code sets min1 = min2 (2), but the remaining stack [5,4,3] has a minimum of 3—so min() returns the wrong value.

• Stack is emptied and reused
  If you push [5], min1 =5 (with no valid min2). Pop the 5 to empty the stack, but min1 still holds 5. Now push [10]—your code won't reset min1 to 10, so min() will incorrectly return 5 instead of the actual minimum 10.

Beyond the failure scenarios, here are the fundamental flaws in the logic:

• No historical tracking of minima
  Storing only two values (min1 and min2) can't capture the full hierarchy of minima as elements are added and removed. When the current min1 is popped, you assume min2 is the new minimum—but this only holds if min2 is still present in the stack and is indeed the next smallest value. This falls apart when min2 was already popped, or when there were multiple instances of min1.

• Unhandled edge cases for small stacks
  Your logic doesn't account for stacks with 0 or 1 elements. For an empty stack, min1 still retains its last value, leading to false results. For a single-element stack, min2 is undefined, which can cause errors if you try to use it during a pop.

• No synchronization for non-minimum pops
  When you pop an element that's neither min1 nor min2, you don't check if that pop affects the min2 value. For example, if min2 is popped, you never update it to the next smallest value in the stack, leaving min2 stale for future operations.

• Ambiguous handling of duplicate minima
  When pushing a value equal to min1, it's unclear how min2 should be updated. If you leave min2 as the previous second smallest, popping one instance of min1 will set min1 to min2—but there are still other instances of min1 in the stack, making the new min1 incorrect.

内容的提问来源于stack exchange，提问作者user5853613