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Copy pathFlatten Nested List Iterator.js
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Flatten Nested List Iterator.js
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/**
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].
*/
/**
* // This is the interface that allows for creating nested lists.
* // You should not implement it, or speculate about its implementation
* function NestedInteger() {
*
* Return true if this NestedInteger holds a single integer, rather than a nested list.
* @return {boolean}
* this.isInteger = function() {
* ...
* };
*
* Return the single integer that this NestedInteger holds, if it holds a single integer
* Return null if this NestedInteger holds a nested list
* @return {integer}
* this.getInteger = function() {
* ...
* };
*
* Return the nested list that this NestedInteger holds, if it holds a nested list
* Return null if this NestedInteger holds a single integer
* @return {NestedInteger[]}
* this.getList = function() {
* ...
* };
* };
*/
/**
* @constructor
* @param {NestedInteger[]} nestedList
*/
var NestedIterator = function(nestedList) {
this.stack = nestedList.reverse();
};
/**
* @this NestedIterator
* @returns {boolean}
*/
NestedIterator.prototype.hasNext = function() {
var next,
i;
while (this.stack.length > 0) {
next = this.stack[this.stack.length - 1];
if (next.isInteger()) {
return true;
}
this.stack.pop();
for (i = next.getList().length - 1; i >= 0; i--) {
this.stack.push(next.getList()[i]);
}
}
return false;
};
/**
* @this NestedIterator
* @returns {integer}
*/
NestedIterator.prototype.next = function() {
return this.stack.pop();
};
/**
* Your NestedIterator will be called like this:
* var i = new NestedIterator(nestedList), a = [];
* while (i.hasNext()) a.push(i.next());
*/