1172. tray stack
We ∞ unlimited number of stacks in a row, from left to right order numbered from 0. Each stack the maximum capacity of the same capacity.
DinnerPlates implement a class called "dish" in:
DinnerPlates (int capacity) - gives the maximum stack capacity capacity.
void push (int val) - val which will be given a positive integer from left to right is pushed into the first stack is not full.
int pop () - returns from right to left top of the stack first non-null value, and remove it from the stack; if all the stacks are empty, return -1.
int popAtStack (int index) - Returns the value of the index number of the top stack and remove it from the stack; index number if the stack is empty, return -1.
Example:
输入：
["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],[],[],[],[],[]]
输出：
[null,null,null,null,null,null,2,null,null,20,21,5,4,3,1,-1]
Explain:
DinnerPlates DinnerPlates = D (2); // initialization, the stack 2 = maximum capacity Capacity
D.push (. 1);
D.push (2);
D.push (. 3);
D.push (. 4);
D. push (5); // stack status is:. 4 2
. 1. 3. 5
﹈ ﹈ ﹈
D.popAtStack (0); // returns 2. Stack status is: 4
. 1. 3. 5
﹈ ﹈ ﹈
D.push (20); // stack status is: 204
. 1. 3. 5
﹈ ﹈ ﹈
D.push (21); // stack status is: 204 21 is
. 1. 3. 5
﹈ ﹈ ﹈
D.popAtStack (0); // returns 20. Situation stack of: 21. 4
. 1. 3. 5
﹈ ﹈ ﹈
D.popAtStack (2); // returns 21. Status stack is: 4
1 3 5
﹈ ﹈ ﹈
D.pop () // returns 5. Status stack is: 4
1 3
﹈ ﹈
D.pop () // returns 4. Status stack is: 13
﹈ ﹈
D.pop () // returns 3. Status stack: 1
﹈
D.pop () // returns 1. Now do not stack.
D.pop () // returns -1. Still do not stack.
prompt:
1 <= Capacity <= 20000
1 <= Val <= 20000
0 <= index <= 100000
most will push, pop, and popAtStack be 200000 times call.
Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/dinner-plate-stacks
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