Lintcode 摊平嵌套的列表

/**
 * // 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:
    vector<int> v;
    int cnt;
    int nested_size;
    NestedIterator(vector<NestedInteger> &nestedList) {
        // Initialize your data structure here.
        v.clear();
        cnt = 0;
        dfs(nestedList);
        nested_size = v.size();
    }

    // @return {int} the next element in the iteration
    int next() {
        // Write your code here
        return v[cnt++];
    }

    // @return {boolean} true if the iteration has more element or false
    bool hasNext() {
        // Write your code here
        if(nested_size){
                nested_size = nested_size - 1;
                return true;
        } else {
                return false;
        }
    }

        void dfs(vector<NestedInteger> nestedList){
                for(int i = 0; i < nestedList.size(); ++i){
                        if(nestedList[i].isInteger()){
                                v.push_back(nestedList[i].getInteger());
                        } else {
                                dfs(nestedList[i].getList());
                        }
                }
        }
};
/**
 * Your NestedIterator object will be instantiated and called as such:
 * NestedIterator i(nestedList);
 * while (i.hasNext()) v.push_back(i.next());
 */