341. Flatten Nested List Iterator

题目：压平嵌套链表迭代器

Given a nested list of integers, implement an iterator to flatten it.

Each element is either an integer, or a list -- whose elements may also be integers or other lists.

Example 1:
Given the list [[1,1],2,[1,1]],

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,1,2,1,1].

Example 2:
Given the list [1,[4,[6]]],

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,4,6].

题意：

给定一个嵌套的整数链表，实现一个迭代器去压平它。

嵌套链表中每一个元素是一个整数或者一个链表--该链表又包含其他整数或者其他链表。

思路一：转载：http://www.cnblogs.com/grandyang/p/5358793.html

使用迭代解，使用栈做辅助遍历，由于栈的后进先出的特性，我们在对向量遍历的时候，从后往前把对象压入栈中，那么第一个对象最后压入栈就会第一个取出来处理，我们的hasNext()函数需要遍历栈，并进行处理，如果栈顶元素是整数，直接返回true，如果不是，那么移除栈顶元素，并开始遍历这个取出的list，还是从后往前压入栈，循环停止条件是栈为空，返回false。

代码：C++版：56ms

/** * // This is the interface that allows for creating nested lists. * // You should not implement it, or speculate about its implementation * class NestedInteger { *   public: *     // Return true if this NestedInteger holds a single integer, rather than a nested list. *     bool isInteger() const; * *     // Return the single integer that this NestedInteger holds, if it holds a single integer *     // The result is undefined if this NestedInteger holds a nested list *     int getInteger() const; * *     // Return the nested list that this NestedInteger holds, if it holds a nested list *     // The result is undefined if this NestedInteger holds a single integer *     const vector<NestedInteger> &getList() const; * }; */class NestedIterator {public:    NestedIterator(vector<NestedInteger> &nestedList) {        for (int i=nestedList.size()-1; i>=0; --i) {  //将nestedList链表反向放入栈中            s.push(nestedList[i]);        }    }    int next() {        NestedInteger t = s.top();        s.pop();        return t.getInteger();    }    bool hasNext() {        while (!s.empty()) {            NestedInteger t = s.top();            if (t.isInteger()) return true; //如果栈顶元素是整数，则返回true            s.pop(); //是链表，则继续讲该链表反向放入栈中            for (int i=t.getList().size()-1; i>=0; --i) {                  s.push(t.getList()[i]);            }        }        return false;    }private:    stack<NestedInteger> s;  //将平整过后的链表放入栈中};/** * Your NestedIterator object will be instantiated and called as such: * NestedIterator i(nestedList); * while (i.hasNext()) cout << i.next(); */

思路二：

使用迭代加双端队列的方式实现。根据双端队列的特点稍微调整代码即可。

代码：C++版：165ms

/** * // This is the interface that allows for creating nested lists. * // You should not implement it, or speculate about its implementation * class NestedInteger { *   public: *     // Return true if this NestedInteger holds a single integer, rather than a nested list. *     bool isInteger() const; * *     // Return the single integer that this NestedInteger holds, if it holds a single integer *     // The result is undefined if this NestedInteger holds a nested list *     int getInteger() const; * *     // Return the nested list that this NestedInteger holds, if it holds a nested list *     // The result is undefined if this NestedInteger holds a single integer *     const vector<NestedInteger> &getList() const; * }; */class NestedIterator {public:    NestedIterator(vector<NestedInteger> &nestedList) {        for (auto a : nestedList) {  //将nestedList链表放入双端队列中            d.push_back(a);        }    }    int next() {        NestedInteger t = d.front();        d.pop_front();        return t.getInteger();    }    bool hasNext() {        while (!d.empty()) {            NestedInteger t = d.front();            if (t.isInteger()) return true; //如果栈顶元素是整数，则返回true            d.pop_front(); //是链表，则继续讲该链表反向放入栈中            for (int i=0; i<t.getList().size(); ++i) {                  d.insert(d.begin()+i, t.getList()[i]);            }        }        return false;    }private:    deque<NestedInteger> d;  //将平整过后的链表放入栈中};/** * Your NestedIterator object will be instantiated and called as such: * NestedIterator i(nestedList); * while (i.hasNext()) cout << i.next(); */

思路三：

虽说迭代器是要用迭代的方法，但是我们可以强行使用递归来解，怎么个强行法呢，就是我们使用一个队列queue，在构造函数的时候就利用迭代的方法把这个嵌套链表全部压平展开，然后在调用hasNext()和next()就很简单了。

代码：C++版：36ms

/** * // This is the interface that allows for creating nested lists. * // You should not implement it, or speculate about its implementation * class NestedInteger { *   public: *     // Return true if this NestedInteger holds a single integer, rather than a nested list. *     bool isInteger() const; * *     // Return the single integer that this NestedInteger holds, if it holds a single integer *     // The result is undefined if this NestedInteger holds a nested list *     int getInteger() const; * *     // Return the nested list that this NestedInteger holds, if it holds a nested list *     // The result is undefined if this NestedInteger holds a single integer *     const vector<NestedInteger> &getList() const; * }; */class NestedIterator {public:    NestedIterator(vector<NestedInteger> &nestedList) {        make_queue(nestedList);    }    int next() {        int t = q.front();        q.pop();        return t;    }    bool hasNext() {        return !q.empty();    }private:    queue<int> q;  //将平整过后的链表放入队列中    void make_queue(vector<NestedInteger> &nestedList) {        for (auto a : nestedList) {            if (a.isInteger()) q.push(a.getInteger());            else make_queue(a.getList());        }    }};/** * Your NestedIterator object will be instantiated and called as such: * NestedIterator i(nestedList); * while (i.hasNext()) cout << i.next(); */