Problem:
Implement FreqStack, a class which simulates the operation of a stack-like data structure.
FreqStack has two functions:
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:
思路:
使用2个哈希表,一个保存频率,一个保存根据频率值确定的元素栈,即每个元素在频率为1到其最大频率的栈内均有值。
Solution (C++):
unordered_map<int, int> freq;
unordered_map<int, stack<int>> stk;
int max_freq = 0;
FreqStack() {
}
void push(int x) {
max_freq = max(max_freq, ++freq[x]);
stk[freq[x]].push(x);
}
int pop() {
if (max_freq == 0) return -1;
int x = stk[max_freq].top();
stk[max_freq].pop();
if (stk[freq[x]--].empty()) --max_freq;
return x;
}
性能:
Runtime: 348 ms??Memory Usage: 72.3 MB
思路:
Solution (C++):
性能:
Runtime: ms??Memory Usage: MB
原文:https://www.cnblogs.com/dysjtu1995/p/12749250.html