标签：constant element empty iat led min() get value tac
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Example:
MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); --> Returns -3. minStack.pop(); minStack.top(); --> Returns 0. minStack.getMin(); --> Returns -2.
public class MinStack {
Stack<Long> stack;
int min;
/** initialize your data structure here. */
public MinStack() {
stack = new Stack<Long>();
min = 0;
}
public void push(int x) {
if (stack.isEmpty()) {
min = x;
stack.push(0L);
return;
}
stack.push((long)x - min);
if (x < min) {
min = x;
}
}
public void pop() {
long cur = stack.pop();
if (cur < 0) {
min = (int)(min - cur);
}
}
public int top() {
if (stack.peek() < 0L) {
return min;
} else {
return (int)(min + stack.peek());
}
}
public int getMin() {
return min;
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
另一种办法
分析：保存到目前为止的最小值。当出栈导致更新最小值时，如果记录了之前的最小值信息，就能更新到ｐｏｐ出当前元素后的最小值。
public class MinStack {
Stack<Integer> stack;
int min;
/** initialize your data structure here. */
public MinStack() {
stack = new Stack<Integer>();
min = Integer.MAX_VALUE;
}
public void push(int x) {
if (x <= min) { //这里要用<=。如果x == min，ｘ也会被判断为新的最小值，在ｐｏｐ时连ｐｏｐ两次，因此此时也要ｐｕｓｈ两次
stack.push(min);
min = x;
}
stack.push(x);
}
public void pop() {
int cur = stack.pop();
if (cur == min) {
min = stack.pop();
}
}
public int top() {
return stack.peek();
}
public int getMin() {
return min;
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
标签：constant element empty iat led min() get value tac
原文地址：http://www.cnblogs.com/yuchenkit/p/7169507.html
