三合一。描述如何只用一个数组来实现三个栈。
你应该实现push(stackNum, value)
、pop(stackNum)
、isEmpty(stackNum)
、peek(stackNum)
方法。stackNum
表示栈下标,value
表示压入的值。
构造函数会传入一个stackSize
参数,代表每个栈的大小。
示例1:
输入:
["TripleInOne", "push", "push", "pop", "pop", "pop", "isEmpty"]
[[1], [0, 1], [0, 2], [0], [0], [0], [0]]
输出:
[null, null, null, 1, -1, -1, true]
说明:当栈为空时`pop, peek`返回-1,当栈满时`push`不压入元素。
示例2:
输入:
["TripleInOne", "push", "push", "push", "pop", "pop", "pop", "peek"]
[[2], [0, 1], [0, 2], [0, 3], [0], [0], [0], [0]]
输出:
[null, null, null, null, 2, 1, -1, -1]
提示:
0 <= stackNum <= 2
//leetcode submit region begin(Prohibit modification and deletion)
class TripleInOne {
private int[] valueArr;
private int[] indexArr;
private int stackSize;
public TripleInOne(int stackSize) {
this.valueArr = new int[stackSize * 3];
this.indexArr = new int[3];
this.indexArr[0] = -1;
this.indexArr[1] = -1;
this.indexArr[2] = -1;
this.stackSize = stackSize;
}
public void push(int stackNum, int value) {
if (this.indexArr[stackNum] == stackSize - 1) {
return;
}
this.indexArr[stackNum]++;
this.valueArr[this.indexArr[stackNum] * 3 + stackNum] = value;
}
public int pop(int stackNum) {
if (this.indexArr[stackNum] == -1) {
return -1;
}
int value = this.valueArr[this.indexArr[stackNum] * 3 + stackNum];
this.indexArr[stackNum]--;
return value;
}
public int peek(int stackNum) {
if (this.indexArr[stackNum] == -1) {
return -1;
}
return this.valueArr[this.indexArr[stackNum] * 3 + stackNum];
}
public boolean isEmpty(int stackNum) {
return this.indexArr[stackNum] == -1;
}
}
/**
* Your TripleInOne object will be instantiated and called as such:
* TripleInOne obj = new TripleInOne(stackSize);
* obj.push(stackNum,value);
* int param_2 = obj.pop(stackNum);
* int param_3 = obj.peek(stackNum);
* boolean param_4 = obj.isEmpty(stackNum);
*/
//leetcode submit region end(Prohibit modification and deletion)