Recover a Quack Data Structure | tech::interview


Recover a Quack Data Structure | tech::interview
给一个Quack的类,里面有三个方法:
pop(): 随机从头或者尾扔出一个元素;
peek(): 随机看头或者尾的一个元素,peek()之后pop()的话一定会pop()出peek()的那个元素;
push(): 向尾部插入一个元素
问题是:给一个排序好的Quack,怎么把里面的元素原封不动的放到一个Array里面。
Follow up:如果quack里面有重复的元素,怎么处理。

对于不重复元素的情况,用queue存小的数,stack存大的数,先pop()一个数,再peek一下,比较这两个数,如果pop的大,就代表肯定是quack的尾巴,反之肯定是头,然后插入queue或者stack就行了。这里假设quack有empty方法。
vector<int> recover_quack(quack& qk) {
queue<int> q;
stack<int> s;
while(!qk.empty()) {
int num1 = qk.pop();
if(qk.empty()) {
q.push(num1);
break;
}
int num2 = qk.peek();
if(num1 < num2) q.push(num1);
else s.push(num1);
}
vector<int> ret;
while(!q.empty()) {
ret.push_back(q.front());
q.pop();
}
while(!s.empty()) {
ret.push_back(s.top());
s.pop();
}
return ret;
}

对于有重复元素的情况,算法跟上面差不多,但是当遇到重复的时候,可以用一个hash_map记下来这个count,最后结算的时候补上去即可。
vector<int> recover_quack(quack& qk) {
queue<int> q;
stack<int> s;
unordered_map<int,int> rep;
while(!qk.empty()) {
int num1 = qk.pop();
if(qk.empty()) {
q.push(num1);
break;
}
int num2 = qk.peek();
if(num1 < num2) q.push(num1);
else if(num1 > num2) s.push(num1);
else {
++rep[num1];
}
}
vector<int> ret;
while(!q.empty()) {
int num = q.front();
ret.push_back(num);
int count = rep[num];
while(count-- > 0) ret.push_back(num);
rep[num] = 0;
q.pop();
}
while(!s.empty()) {
int num = s.top();
ret.push_back(num);
int count = rep[num];
while(count-- > 1) ret.push_back(num);
rep[num] = 0;
s.pop();
}
return ret;
}

参考的quack类,测试用
quack(const vector<int>& init) {
for(auto i : init) {
_data.push_back(i);
}
}
bool empty() {
return _data.empty();
}
int pop() {
int ret = 0;
int ran = _last_peek == -1 ? rand() % 2 : _last_peek;
if(ran == 0) {
ret = _data.front();
_data.pop_front();
} else {
ret = _data.back();
_data.pop_back();
}
_last_peek = -1;
return ret;
}
void push(int num) {
_data.push_back(num);
}
int peek() {
int ret = 0;
if(rand() % 2 == 0) {
ret = _data.front();
_last_peek = 0;
} else {
ret = _data.back();
_last_peek = 1;
}
return ret;
}
private:
deque<int> _data;
int _last_peek{ -1 };
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