Loading

Leetcode每日一题 341. 扁平化嵌套列表迭代器

341. 扁平化嵌套列表迭代器

给你一个嵌套的整型列表。请你设计一个迭代器,使其能够遍历这个整型列表中的所有整数。

列表中的每一项或者为一个整数,或者是另一个列表。其中列表的元素也可能是整数或是其他列表。

 

示例 1:

输入: [[1,1],2,[1,1]]
输出: [1,1,2,1,1]


解释: 通过重复调用 next 直到 hasNext 返回 false,next 返回的元素的顺序应该是: [1,1,2,1,1]。
示例 2:

输入: [1,[4,[6]]]
输出: [1,4,6]


解释: 通过重复调用 next 直到 hasNext 返回 false,next 返回的元素的顺序应该是: [1,4,6]。

 

递归就行了。

 

/**
 * // 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 {
private:
    vector<int> v;
    vector<int>::iterator cur;

    void dfs(const vector<NestedInteger> &nestedList) {
        for (auto &nest : nestedList) {
            if (nest.isInteger()) {
                v.push_back(nest.getInteger());
            } else {
                dfs(nest.getList());
            }
        }
    }

public:
    NestedIterator(vector<NestedInteger> &nestedList) {
        dfs(nestedList);
        cur = v.begin();
    }

    int next() {
        return *cur++;
    }

    bool hasNext() {
        return cur != v.end();
    }
};


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

 

posted @ 2021-03-26 22:57  eveilcoo  阅读(59)  评论(0编辑  收藏  举报