【LeetCode341】-扁平化嵌套列表迭代器

实现思路：

利用递归的特性，如果返回的还是列表就遍历，直到返回的是一个整数，就将该整数压入到队列中，在实现next和hasNext操作的过程中，next操作直接返回队首并弹出，hasNext直接利用队列的特性，判断队列是否为空即可

实现代码：

#include <iostream>
#include <queue>
#include <vector>
using namespace std;


 // 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:
	queue<int> re;
	void push_data(vector<NestedInteger> nestedList) {
    
    
		int n = nestedList.size();
		for (int i = 0;i < n;i++) {
    
    
			if (nestedList[i].isInteger()) re.push(nestedList[i].getInteger());
			else {
    
    
				push_data(nestedList[i].getList());
			}
		}
	}
	NestedIterator(vector<NestedInteger>& nestedList) {
    
    
		push_data(nestedList);
	}

	int next() {
    
    
		int t = re.front();
		re.pop();
	}

	bool hasNext() {
    
    
		return !re.empty();
	}
};

/**
 * Your NestedIterator object will be instantiated and called as such:
 * NestedIterator i(nestedList);
 * while (i.hasNext()) cout << i.next();
 */
int main() {
    
    
	return 0;
}

提交结果及分析：

next操作和hasNext操作的时间复杂度都是O（1），空间复杂度是O（n）

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