341. Flatten Nested List Iterator

问题描述

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].

题目链接：

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思路分析

给一组嵌套的string的向量，要求实现一个类，能够判断list中是否还有数据以及返回下一个数。因为是嵌套在list当中的，下一个数据不一定是int的。

问题的关键在于，如何存储这个Nested List，以及如何解构让它的下一个值是int。第一个问题，使用栈来存储，让最后一个元素先入栈，保持顺序一致。第二个问题，我们要保证栈顶元素是int，这样就可以简单的返回值，实现next方法。

然后在hasNext方法中，实现问题的结构，当栈不为空时进行循环，当栈顶元素是int时可以返回true了，如果不是int的话，那需要继续结构，将list内的元素再次入栈。当栈空了，说明没有元素了，就可以返回false。

代码

/**
 * // 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> stk;
    NestedIterator(vector<NestedInteger> &nestedList) {
        for(int i = nestedList.size() - 1; i >= 0; i--){
            stk.push(nestedList[i]);
        }
    }

    int next() {
        int num = stk.top().getInteger();
        stk.pop();
        return num;
    }

    bool hasNext() {
        while(!stk.empty()){
            if(stk.top().isInteger()){
                return true;
            }
            else{
                NestedInteger temp = stk.top();
                stk.pop();
                vector<NestedInteger> list = temp.getList();
                for(int i = list.size() - 1; i >= 0; i--){
                    stk.push(list[i]);
                }
            }
        }
        return false;
    }
};

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

时间复杂度：$O(n)$ //list中int的个数
空间复杂度：$O(n)$

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反思