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LeetCode 解题报告(345) -- 341. Flatten Nested List Iterator

Problem

You are given a nested list of integers nestedList. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.

Implement the NestedIterator class:

  • NestedIterator (List<NestedInteger> nestedList) Initializes the iterator with the nested list nestedList.
  • int next () Returns the next integer in the nested list.
  • boolean hasNext () Returns true if there are still some integers in the nested list and false otherwise.

Example 1:

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Input: nestedList = [[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:

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Input: nestedList = [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].

Constraints:

  • 1 <= nestedList.length <= 500
  • The values of the integers in the nested list is in the range [-106, 106].

Analysis

   题目给出了一个嵌套链表,要求我们实现一个迭代器去遍历这个嵌套链表。链表中的元素可以是数字,也可以是链表,每个元素的类型提供接口判断。

   一开始的想法是维护一个下标,指向当前已经遍历到那个位置了,这个思路对普通链表来说是没问题的,但是对于嵌套链表,无法快速定位到当前指向的位置。既然在使用过程中我们无法做到,干脆初始化的时候把嵌套链表打平就好,然后按照数组一个一个往后读。

   所以这道题目就转变为打平一个嵌套链表,其实就是 DFS 了。把元素都提取出来放到一个数组中,然后维护一个下标表示当前访问到哪个位置即可。


Solution

   无


Code

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/**
* // 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:
NestedIterator (vector<NestedInteger> &nestedList) {
cursor = 0;
int size = nestedList.size ();
for (int i = 0; i < size; i++) {
helper (arr, nestedList [i]);
}

for (int i = 0; i < arr.size (); i++) {
cout << arr [i] << " ";
}
cout << endl;
}

int next() {
return arr [cursor++];
}

bool hasNext() {
return cursor < arr.size ();
}
private:
vector<int> arr;
int cursor;
void helper(vector<int> &result, NestedInteger integer) {
if (integer.isInteger ()) {
result.push_back (integer.getInteger ());
} else {
vector<NestedInteger> list = integer.getList ();
int size = list.size ();
for (int i = 0; i < size; i++) {
helper (result, list[i]);
}
}
//return result;
}
};

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

Summary

   这道题目难度不大,是数组中一个简单的 DFS。这道题目的分享到这里,感谢你的支持!

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