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341. Flatten Nested List Iterator

2016-05-11 09:07 337 查看
Given a nested list of integers, implement an iterator to flatten it.

Each element is either an integer, or a list -- whose elements may also be integers or other lists.

Example 1:

Given the list
[[1,1],2,[1,1]]
,

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be:
[1,1,2,1,1]
.

Example 2:

Given the list
[1,[4,[6]]]
,

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be:
[1,4,6]
.

【思路】建一个栈,从list尾巴倒序压入栈中,每次检测栈顶,如果是整数元素,直接弹出;如果是向量,则将整个向量先弹出,再倒序压入每一个向量中的元素。

/**
* // 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) {
for(int i = nestedList.size()-1; i>=0; i--)
{
stk.push(nestedList[i]);
}
}

int next() {
int val;
if(hasNext())
{
val = stk.top().getInteger();
stk.pop();
}
return val;
}

bool hasNext() {
while(!stk.empty())
{
NestedInteger tmp = stk.top();
if(tmp.isInteger())
return true;
else{
vector<NestedInteger> tmpList = stk.top().getList();
stk.pop();
for(int j = tmpList.size()-1; j>=0; j--)
{
stk.push(tmpList[j]);
}
}
}
return false;
}
private:
stack<NestedInteger>stk;
};

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