题目
Implement FreqStack, a class which simulates the operation of a stack-like data structure.
FreqStack has two functions:
push(int x), which pushes an integer x onto the stack.
pop(), which removes and returns the most frequent element in the stack.
If there is a tie for most frequent element, the element closest to the top of the stack is removed and returned.
Example 1:
Input:
["FreqStack","push","push","push","push","push","push","pop","pop","pop","pop"],
[[],[5],[7],[5],[7],[4],[5],[],[],[],[]]
Output: [null,null,null,null,null,null,null,5,7,5,4]
Explanation:
After making six .push operations, the stack is [5,7,5,7,4,5] from bottom to top. Then:
pop() -> returns 5, as 5 is the most frequent.
The stack becomes [5,7,5,7,4].
pop() -> returns 7, as 5 and 7 is the most frequent, but 7 is closest to the top.
The stack becomes [5,7,5,4].
pop() -> returns 5.
The stack becomes [5,7,4].
pop() -> returns 4.
The stack becomes [5,7].
Note:
- Calls to FreqStack.push(int x) will be such that 0 <= x <= 10^9.
- It is guaranteed that FreqStack.pop() won't be called if the stack has zero elements.
- The total number of FreqStack.push calls will not exceed 10000 in a single test case.
- The total number of FreqStack.pop calls will not exceed 10000 in a single test case.
- The total number of FreqStack.push and FreqStack.pop calls will not exceed 150000 across all test cases.
题目大意
实现 FreqStack,模拟类似栈的数据结构的操作的一个类。
FreqStack 有两个函数:
- push(int x),将整数 x 推入栈中。
- pop(),它移除并返回栈中出现最频繁的元素。如果最频繁的元素不只一个,则移除并返回最接近栈顶的元素。
解题思路
FreqStack 里面保存频次的 map 和相同频次 group 的 map。push 的时候动态的维护 x 的频次,并更新到对应频次的 group 中。pop 的时候对应减少频次字典里面的频次,并更新到对应频次的 group 中。