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.

题意：实现一个最小栈，要求具有以上四个功能，且能常量时间返回检索最小值元素的功能。

思路：一个数组用来存值，一个数存最小值。注意：数组里存的是存入值与最小值的差，所以可能越界，要用long long类型保存差值。

class MinStack {
public:
void push(int x) {
if (val.empty()){
val.push_back(0);
min = x;
}
else{
val.push_back((long long)x - min);
if (x < min)
min = x;
}
}
void pop() {
if (!val.empty()){
int index = val.size() - 1;
if (val[index] < 0){
min -= val[index];
val.erase(val.begin()+index);
}
else{
val.erase(val.begin() + index);
}
}
}
int top() {
int index = val.size() - 1;
if (val[index] < 0)
return min;
else{
return min + val[index];
}
}
int getMin() {
if (!val.empty())
return min;
else
return -1;
}
private:
vector<long long> val;
int min;
};