341. Flatten Nested List Iterator

https://leetcode.com/problems/flatten-nested-list-iterator/description/

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 callingnextrepeatedly untilhasNextreturns false, the order of elements returned bynextshould be:[1,1,2,1,1].

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

By callingnextrepeatedly untilhasNextreturns false, the order of elements returned bynextshould be:[1,4,6].

Thoughts

实现针对nested的list的iterator，效果和遍历展开了的list一样。处理nested用stack。不过做这题时思路被另一道二维list iterator解法带偏了，总想着要用指针。实际并不需要，倒着把所有元素都压入栈，hasNext遍历栈顶时当遇到nested的元素则把它展开后重新入栈。

Code

/*
 * @lc app=leetcode id=341 lang=cpp
 *
 * [341] Flatten Nested List Iterator
 */

// @lc code=start
/**
 * // 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:
    stack<NestedInteger> s;
    NestedIterator(vector<NestedInteger> &nestedList) {
        for (int i = nestedList.size() - 1; i >= 0; --i) {
            s.push(nestedList[i]);
        }
    }

    int next() {
        int r = s.top().getInteger();
        s.pop();
        return r;
    }

    bool hasNext() {
        while (!s.empty()) {
            if (s.top().isInteger()) return true;
            const auto t = s.top().getList(); s.pop();
            for (int i = t.size() - 1; i >= 0; --i) {
                s.push(t[i]);
            }
        }
        return false;
    }
};

/**
 * Your NestedIterator object will be instantiated and called as such:
 * NestedIterator i(nestedList);
 * while (i.hasNext()) cout << i.next();
 */
// @lc code=end

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * public interface NestedInteger {
 *
 *     // @return true if this NestedInteger holds a single integer, rather than a nested list.
 *     public boolean isInteger();
 *
 *     // @return the single integer that this NestedInteger holds, if it holds a single integer
 *     // Return null if this NestedInteger holds a nested list
 *     public Integer getInteger();
 *
 *     // @return the nested list that this NestedInteger holds, if it holds a nested list
 *     // Return null if this NestedInteger holds a single integer
 *     public List<NestedInteger> getList();
 * }
 */
public class NestedIterator implements Iterator<Integer> {
    Stack<NestedInteger> stack;
    public NestedIterator(List<NestedInteger> nestedList) {
        stack = new Stack();
        for (int i = nestedList.size() - 1; i >= 0; i--) {
            stack.push(nestedList.get(i));
        }
    }

    @Override
    public Integer next() {
        return stack.pop().getInteger();
    }

    @Override
    public boolean hasNext() {
        while (!stack.isEmpty()) {
            if (stack.peek().isInteger()) {
                return true;
            }
            List<NestedInteger> nestedList = stack.pop().getList();
            for (int i = nestedList.size() - 1; i >= 0; i--) {
                stack.push(nestedList.get(i));
            }
        }

        return false;
    }
}

/**
 * Your NestedIterator object will be instantiated and called as such:
 * NestedIterator i = new NestedIterator(nestedList);
 * while (i.hasNext()) v[f()] = i.next();
 */

Analysis

每个元素进出一次栈，时空复杂度O(N).