题目描述:
设计一个支持 push ,pop ,top 操作,并能在常数时间内检索到最小元素的栈。
push(x) —— 将元素 x 推入栈中。
pop() —— 删除栈顶的元素。
top() —— 获取栈顶元素。
getMin() —— 检索栈中的最小元素。
示例:
输入:
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
输出:
[null,null,null,null,-3,null,0,-2]
解释:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.getMin(); --> 返回 -2.
Java代码:
class MinStack {
private int[] arr = null;
private int min;
private int n = 0;
/** initialize your data structure here. */
public MinStack() {
arr = new int[4];
min = Integer.MAX_VALUE;
}
public void push(int x) {
if(n == arr.length) resize(2 * n);
arr[n++] = x;
if(min > x) min = x;
}
public void pop() {
if(arr.length > 4 && n <= arr.length / 4) resize(arr.length / 2);
if(arr[n - 1] == min) {
int temp = Integer.MAX_VALUE;
for(int i = 0;i < n- 1;i++) {
if(arr[i] < temp) temp = arr[i];
}
min = temp;
}
n--;
}
public int top() {
return arr[n - 1];
}
public int getMin() {
return min;
}
private void resize(int newSize) {
int[] temp = new int[newSize];
for(int i = 0;i < arr.length;i++) {
temp[i] = arr[i];
}
this.arr = temp;
}
}
/**
* 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();
*/
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