[LeetCode] 1172. Dinner Plate Stacks
grandyang opened this issue · comments
You have an infinite number of stacks arranged in a row and numbered (left to right) from 0
, each of the stacks has the same maximum capacity.
Implement the DinnerPlates
class:
DinnerPlates(int capacity)
Initializes the object with the maximum capacity of the stackscapacity
.void push(int val)
Pushes the given integerval
into the leftmost stack with a size less thancapacity
.int pop()
Returns the value at the top of the rightmost non-empty stack and removes it from that stack, and returns-1
if all the stacks are empty.int popAtStack(int index)
Returns the value at the top of the stack with the given indexindex
and removes it from that stack or returns-1
if the stack with that given index is empty.
Example 1:
Input
["DinnerPlates", "push", "push", "push", "push", "push", "popAtStack", "push", "push", "popAtStack", "popAtStack", "pop", "pop", "pop", "pop", "pop"]
[[2], [1], [2], [3], [4], [5], [0], [20], [21], [0], [2], [], [], [], [], []]
Output
[null, null, null, null, null, null, 2, null, null, 20, 21, 5, 4, 3, 1, -1]
Explanation:
DinnerPlates D = DinnerPlates(2); // Initialize with capacity = 2
D.push(1);
D.push(2);
D.push(3);
D.push(4);
D.push(5); // The stacks are now: 2 4
1 3 5
﹈ ﹈ ﹈
D.popAtStack(0); // Returns 2. The stacks are now: 4
1 3 5
﹈ ﹈ ﹈
D.push(20); // The stacks are now: 20 4
1 3 5
﹈ ﹈ ﹈
D.push(21); // The stacks are now: 20 4 21
1 3 5
﹈ ﹈ ﹈
D.popAtStack(0); // Returns 20. The stacks are now: 4 21
1 3 5
﹈ ﹈ ﹈
D.popAtStack(2); // Returns 21. The stacks are now: 4
1 3 5
﹈ ﹈ ﹈
D.pop() // Returns 5. The stacks are now: 4
1 3
﹈ ﹈
D.pop() // Returns 4. The stacks are now: 1 3
﹈ ﹈
D.pop() // Returns 3. The stacks are now: 1
﹈
D.pop() // Returns 1. There are no stacks.
D.pop() // Returns -1. There are still no stacks.
Constraints:
1 <= capacity <= 2 * 104
1 <= val <= 2 * 104
0 <= index <= 105
- At most
2 * 105
calls will be made topush
,pop
, andpopAtStack
.
这道题定义了一种餐盘栈的数据结构,说是有很多个栈从左到右排成一排,每个栈有个最大容量 capacity,现在定义了三种操作,push 是将给定的数字 val 压入左起第一个未满的栈,pop 是移除并返回右起第一个非空的栈顶元素,不存在则返回 -1。popAtStack 是移除并返回给定 index 位置的栈顶元素,不存在则返回 -1。题目的例子给了详细的解释,不难理解题意,但一定要将屏幕拉宽,让栈中的数字对齐,不然会产生疑惑,至少博主第一次看的时候产生了。既然是要有很多栈,那么可以把这些栈都放到一个数组中,栈本身可以用个 vector 来代替,所以这里就用个二维数组 stacks 就可以了。然后再来想这道题的难点是什么,对于 push 函数来说,要找到左起第一个未满的栈,怎么才能知道哪个栈未满呢,一个一个去遍历吗?未免太不高效了,最好这里能记录所有未满的栈的位置,所以这里用个 TreeSet 来记录,可以利用其自动排序的特点,则第一个元素一定是左起第一个栈,最后一个元素一定是右起第一个栈。首先来判定 openSt 是否为空,若为空,便是当前没有未满的栈,则需要新增一个空栈,将该新栈位置加入 openSt 中,并且 stacks 扩大一位。然后取出 openSt 的首元素,即为左起第一个未满的栈,将 val 加入其中。然后需要判断加入后该栈是否已经满了,是的话则从 openSt 中移除该位置。对于 pop 来说,这里可以直接调用 popAtStack 函数,传入 stacks 的末尾位置。对于 popAtStack 来说,首先要判断给定的 index 参数是否越界,若越界或者对应的栈为空,则直接返回 -1。否则取出对应的栈顶元素,然后将当前位置 index 加入 openSt,因为此时栈未满了。接下来有个很重要的操作,从右起移除所有为空的栈,不然下次调用 pop 后可能无法返回正确元素。用一个 while 循环,若右起第一个栈为空,则移除,同时将 openSt 中对应的位置也要移除,参见代码如下:
class DinnerPlates {
public:
DinnerPlates(int capacity) {
cap = capacity;
}
void push(int val) {
if (openSt.empty()) {
openSt.insert(stacks.size());
stacks.resize(stacks.size() + 1);
}
stacks[*openSt.begin()].push_back(val);
if (stacks[*openSt.begin()].size() == cap) {
openSt.erase(openSt.begin());
}
}
int pop() {
return popAtStack((int)stacks.size() - 1);
}
int popAtStack(int index) {
if (index < 0 || index >= stacks.size() || stacks[index].empty()) {
return -1;
}
int res = stacks[index].back();
stacks[index].pop_back();
openSt.insert(index);
while (!stacks.empty() && stacks.back().empty()) {
stacks.pop_back();
openSt.erase(*openSt.rbegin());
}
return res;
}
private:
vector<vector<int>> stacks;
set<int> openSt;
int cap;
};
Github 同步地址:
参考资料:
https://leetcode.com/problems/dinner-plate-stacks/
https://leetcode.com/problems/dinner-plate-stacks/discuss/366331/C%2B%2BPython-Two-Solutions