leetcode 155 Min Stack
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.
解决方案：
可以看到STL的解决方案跟大多数的c语言解决方案还是有差距的，后序我找一找基于链表的整齐点的c语言实现
The key idea is use a another stack to store the minimum value of the corresponding stack. Put differently, min[i] equals the minimum element where data[i] is the top of this sub-stack.
We can use a full size of min where it’s size equals the data’s, but it’s not necessary.
I have 2 main concerns about the algorithm:
1 We should pop the element in min IFF there’s match of data.top().
2 If we have multiple minima, for example [0, 1, 0] in data, then the min should be [0, 0]. Otherwise, the the pop operation wouldn’t work properly. As a result, we should push the element if x <= min.top().
class MinStack {
public:
void push(int x) {
s.push(x);
if (mins.empty() || x<=mins.top()) {
mins.push(x);
}
}
void pop() {
int temp = s.top();
s.pop();
if (temp == mins.top()) {
mins.pop();
}
}
int top() {
return s.top();
}
int getMin() {
return mins.top();
}
private:
stack<int> s;
stack<int> mins;
};STL list实现：
class MinStack {
private:
list<int> s;
int min;
public:
MinStack()
{
min=INT_MAX;
}
void push(int x) {
if(x<min) min=x;
s.push_back(x);
}
void pop() {
if(s.back()==min)
{
s.pop_back();
min=INT_MAX;
list<int>::iterator it=s.begin();
while(it!=s.end())
{
if(*it<min) min=*it;
it++;
}
}else
s.pop_back();
}
int top() {
return s.back();
}
int getMin() {
return min;
}
};python解决方案：
class MinStack:
# @param x, an integer
def __init__(self):
# the stack it self
self.A = []
self.minS=[]
# @return an integer
def push(self, x):
n=len(self.A)
if n==0:
self.minS.append(x)
else:
lastmin=self.minS[-1]
if x<=lastmin:
self.minS.append(x)
self.A.append(x)
# @return nothing
def pop(self):
if len(self.A)>0 and self.A.pop()==self.minS[-1]:
self.minS.pop()
# @return an integer
def top(self):
return self.A[-1]
# @return an integer
def getMin(self):
return self.minS[-1]python解决方案2：
class MinStack:
def __init__(self):
self.q = []
# @param x, an integer
# @return an integer
def push(self, x):
curMin = self.getMin()
if curMin == None or x < curMin:
curMin = x
self.q.append((x, curMin));
# @return nothing
def pop(self):
self.q.pop()
# @return an integer
def top(self):
if len(self.q) == 0:
return None
else:
return self.q[len(self.q) - 1][0]
# @return an integer
def getMin(self):
if len(self.q) == 0:
return None
else:
return self.q[len(self.q) - 1][1]
asked Apr 14 in Min Stack by charles8135 (180 points) 腾讯云开发者
