Problem:
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
- push(x) -- Push element x onto stack.
- pop() -- Removes the element on top of the stack.
- top() -- Get the top element.
- getMin() -- Retrieve the minimum element in the stack.
Example 1:
Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
Output
[null,null,null,null,-3,null,0,-2]
Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top(); // return 0
minStack.getMin(); // return -2
Constraints:
- Methods pop, top and getMin operations will always be called on non-empty stacks.
思路：
定义一个栈与保存最小值的栈。最小值栈与栈同时push和pop，添加值的时候，与最小值栈顶元素比较，若新值小则将新值压入最小值栈，否则将最小值栈顶的元素再次压栈。
Solution (C++):
stack<int> stk, min_stk;
/** initialize your data structure here. */
MinStack() {
}
void push(int x) {
stk.push(x);
if (min_stk.empty() || x < min_stk.top())
min_stk.push(x);
else min_stk.push(min_stk.top());
}
void pop() {
if (!stk.empty() && !min_stk.empty()) {
stk.pop();
min_stk.pop();
}
}
int top() {
if (stk.empty()) return -1;
return stk.top();
}
int getMin() {
if (min_stk.empty()) return -1;
return min_stk.top();
}
性能：
Runtime: 40 ms Memory Usage: 14.5 MB
思路：
也是定义一个data栈与保存最小值的栈，但是与前一种方法不同，入栈时检查新值小于等于最小值栈顶的值，如果是则压入最小值栈，否则不压入最小值栈。pop时判断data栈顶值是否等于最小值栈顶值，若是则最小值栈顶弹出，否则不弹出。
Solution (C++):
stack<int> stk, min_stk;
/** initialize your data structure here. */
MinStack() {
}
void push(int x) {
stk.push(x);
if (min_stk.empty() || x <= min_stk.top())
min_stk.push(x);
}
void pop() {
if (!stk.empty()) {
if (stk.top() == min_stk.top()) min_stk.pop();
stk.pop();
}
}
int top() {
if (stk.empty()) return -1;
return stk.top();
}
int getMin() {
if (min_stk.empty()) return -1;
return min_stk.top();
}
性能：
Runtime: 44 ms Memory Usage: 14.3 MB
