給定一個嵌套的整型列表。設計一個迭代器,使其能夠遍歷這個整型列表中的所有整數。
列表中的項或者爲一個整數,或者是另一個列表。
示例 1:
輸入: [[1,1],2,[1,1]]
輸出: [1,1,2,1,1]
解釋: 通過重複調用 next 直到 hasNext 返回false,next 返回的元素的順序應該是: [1,1,2,1,1]。
示例 2:
輸入: [1,[4,[6]]]
輸出: [1,4,6]
解釋: 通過重複調用 next 直到 hasNext 返回false,next 返回的元素的順序應該是: [1,4,6]。
解題思路:
首先要仔細讀懂題目及其所給的信息,當你看明白之後,其實這個題也就差不多了。大概思路是:從頭開始遍歷nestedList,若是Int直接Push進Res中,若不是,那就遞歸。
代碼如下:
/**
* // 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) {
res(nestedList);
}
void res(vector<NestedInteger> &nestedList){
for(auto t:nestedList){
if(t.isInteger()){
ans.push_back(t.getInteger());
}
else{
res(t.getList());
}
}
}
int next() {
return ans[cnt++];
}
bool hasNext() {
return cnt<ans.size();
}
private:
vector<int> ans;
int cnt=0;
};
/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i(nestedList);
* while (i.hasNext()) cout << i.next();
*/