leetcode 341. Flatten Nested List Iterator

妖狐艹你老母 2022-06-07 01:14 152阅读 0赞

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: \[1,4,6\].

嗯嗯，这道题先这样把，不喜欢这样的题。

代码如下：

/**
     * // 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<>();
            pushData(nestedList);
        }
    
        @Override
        public Integer next() {
            return stack.pop().getInteger();
        }
    
        @Override
        public boolean hasNext() {
            while(!stack.isEmpty()) {
                if (stack.peek().isInteger()) {
                    return true;
                }
                pushData(stack.pop().getList());
            }
            return false;
        }
    
        private void pushData(List<NestedInteger> nestedList) {
            for (int i = nestedList.size() - 1; i >= 0; i--) {
                stack.push(nestedList.get(i));
            }
        }
    }
    
    /**
     * Your NestedIterator object will be instantiated and called as such:
     * NestedIterator i = new NestedIterator(nestedList);
     * while (i.hasNext()) v[f()] = i.next();
     */

下面是C++的做法，

这道题让我们建立压平嵌套链表的迭代器，关于嵌套链表的数据结构最早出现在Nested List Weight Sum中，而那道题是用的递归的方法来解的，而迭代器一般都是用迭代的方法来解的，而递归一般都需用栈来辅助遍历，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，

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    
    class NestedIterator 
    {
    public:
        NestedIterator(vector<NestedInteger> &nestedList)
        {
            for (int i = nestedList.size() - 1; i >= 0; --i) 
            {
                s.push(nestedList[i]);
            }
        }
    
        int next() 
        {
            NestedInteger t = s.top(); s.pop();
            return t.getInteger();
        }
    
        bool hasNext() 
        {
            while (s.empty()==false) 
            {
                NestedInteger t = s.top();
                if (t.isInteger()) return true;
                s.pop();
                for (int i = t.getList().size() - 1; i >= 0; --i) 
                {
                    s.push(t.getList()[i]);
                }
            }
            return false;
        }
    
    private:
        stack<NestedInteger> s;
    };