341 Flatten Nested List Iterator

341. 1. Question

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].

2. Implementation

思路：利用stack，从输入的Nested Integer List的后面逐个将List中的元素放到stack上

(1) hasNext(): 查看栈顶的元素是否Integer, 是的话直接返回true，否则栈顶元素代表一个list，pop出它，并展开这个list，从后面逐个将每个元素放在stack上

(2) next(): 用hasNext()判断是否还有下一个元素，如果有的话pop出stack的元素并且call getInteger()，否则return null

/**
 * // 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 = null;

    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()) {
            NestedInteger curNI = stack.peek();
            if (curNI.isInteger()) {
                return true;
            }
            stack.pop();

            for (int i = curNI.getList().size() - 1; i >= 0; i--) {
                stack.push(curNI.getList().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();
 */

3. Time & Space Complexity

时间和空间都是O(n)

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