[LeetCode] 341. Flatten Nested List Iterator

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:

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

Example 2:

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

扁平化嵌套列表迭代器。

题意是给一个带有嵌套的列表，里面存的是整数。请你把它扁平化，满足当调用hasNext()函数的时候，能返回相应的数字。我这里提供两种做法，一种是递归，另一种用到stack。

首先是递归的做法。首先创建一个空的列表res和一个index变量，记录list遍历到的位置。因为是嵌套列表，所以当遍历input列表的时候，有可能是数字，也有可能不是。如果不是数字，则继续递归调用helper函数，去处理sub list。如果是数字，则把数字加入res列表。其他的部分就比较直观了。

时间O(n)

空间O(n)

Java实现

public class NestedIterator implements Iterator<Integer> {
    List<Integer> list = new ArrayList<>(); // 普通List
    int index = 0;

    public NestedIterator(List<NestedInteger> nestedList) {
        // 预处理，扁平化嵌套list
        helper(nestedList);
    }

    public void helper(List<NestedInteger> nestedList) {
        // 循环List中每一个对象
        for (NestedInteger i : nestedList) {
            // 如果是数字
            if (i.isInteger()) {
                // 将该数字加入普通List
                list.add(i.getInteger());
            } else {
                // 如果不是数字，递归子问题解决
                helper(i.getList());
            }
        }
    }

    @Override
    public Integer next() {
        if (!hasNext()) {
            return null;
        }
        return list.get(index++);
    }

    @Override
    public boolean hasNext() {
        return index < list.size();
    }
}

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

另一种做法是用到栈stack。既然用到stack了，而且stack是先进后出，所以试着从input list的尾部往头部遍历。如果遍历到的是数字integer，则把这个数字加入stack；如果遍历到的不是integer，那么在把这个list从stack里弹出的同时，也从其尾部往头部遍历，将遍历到的数字加入stack。

时间O(n)

空间O(n)

Java实现

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()) {
            NestedInteger cur = stack.peek();
            if (cur.isInteger()) {
                return true;
            }
            stack.pop();
            for (int i = cur.getList().size() - 1; i >= 0; i--) {
                stack.push(cur.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();
 */

LeetCode 题目总结