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].

题意:为一个 nestedList 实现一个 iterator。

解法一. 直接把 nestedlist 展开成一个 list. 代码

/**
 * // 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> {
    List<Integer> list = new ArrayList<>();
    int index = 0;

    public void helper(List<NestedInteger> nestedList, List<Integer> list, int index) {
        if(index >= nestedList.size()) return;
        NestedInteger cur = nestedList.get(index);
        if(cur.isInteger()) {
            list.add(cur.getInteger());
        } else {
            helper(cur.getList(), list, 0);
        }
        helper(nestedList, list, index + 1);
    }
    public NestedIterator(List<NestedInteger> nestedList) {
        helper(nestedList, list, 0);
        for(int i : list) {
            System.out.println(i + " ");
        }
    }

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

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

解法二:使用 stack。在构造函数里,我们从后往前把值 push 到栈里。 当调用 hasNext func 的时候,如果 pop 出来是 list 的话,我们把 list 展开从后 push 到栈里。调用 next func, 我们直接返回栈顶。注意 hasNext func 里面的 while 是为了处理空 list 的情况。

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() {
        if(stack.isEmpty()) return false;
        while(!stack.isEmpty()) {
            NestedInteger cur = stack.pop();
            if(cur.isInteger())  {
                stack.push(cur);
                return true;
            } else {
                for(int i = cur.getList().size() - 1; i >= 0; i--) {
                    stack.push(cur.getList().get(i));
                }
            }
        }
        return !stack.isEmpty();
    }
}

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