leetcode 341. Flatten Nested List Iterator

本题的难点是，这不是一个双重数组，数组里面是一个NestedInteger类，它有isInteger, getInteger, getList等方法。因此很容易被示例中的 [[1,1],2,[1,1]]所误导。

 class NestedIterator {
      constructor(nestedList) {
        this.s = nestedList
      }
      next() {
        var el = this.s.shift()
        return el.getInteger()
      }

      hasNext() {

        var s = this.s
        while (s.length) {
          var temp = s[0]
          if (temp.isInteger()) {
            return true;
          }
          s.shift();
          //如果temp为一个数组
          for (var i = temp.getList().length - 1; i >= 0; i--) {
            s.unshift(temp.getList()[i]);
          }
        }
        return false;
      }
    }

如果传参真的是一个 [[1,1],2,[1,1]]数组又如何呢，并且它支持如下两种遍历。

//遍历1
    var i = new NestedIterator([[1, 2, 2.5], [], 3, [4, 5]]), a = [];

    do {
      var val = i.next();
      if (typeof val == 'number') {
        a.push(val)
      } else {
        break
      }
    } while (true)


    console.log(a)

======

//遍历2
    var i = new NestedIterator([[1, 2, 2.5], [], 3, [4, 5]]), a = [];
    while (i.hasNext()) a.push(i.next());
    console.log(a)

这时我们需要引进一个变量，让它无论如何 都走一下hasNext方法

    class NestedIterator {
      constructor(list) {
        this.s = list
        this._hasCallHasNext = false;
      }
      next() {
        if (!this._hasCallHasNext) {
          this.hasNext()
        }
        var el = this.s.shift()
        this._hasCallHasNext = false;
        return el
      }

      hasNext() {

        var s = this.s
        while (s.length) {
          var temp = s[0]
          if (typeof temp === 'number') {
            this._hasCallHasNext = true;
            return true;
          }
          s.shift();
          //如果temp为一个数组
          for (var i = temp.length - 1; i >= 0; i--) {
            s.unshift(temp[i]);
          }
        }
        return false;
      }
    }

但这会改变数组的结构，更普适的方案是添加两个游标：

 class NestedIterator {
      constructor(list) {
        this.s = list
        this.I = 0;
        this.J = 0;
        this._hasCallHasNext = false;
      }
      next() {
        if (!this._hasCallHasNext) {
          this.hasNext()
        }
        var list = this.s[this.I], val
        if (Array.isArray(list)) {
          val = list[this.J]
          this.J++;
        } else {
          val = list;
          this.I++
          this.J = 0;
        }
        this._hasCallHasNext = false
        return val
      }
      hasNext() {
        var s = this.s;
        for (var i = this.I; i < s.length; i++) {
          var temp = s[i]
          if (typeof temp === 'number') {
            this.I = i
            this._hasCallHasNext = true;
            return true;
          }
          if (this.J > temp.length) {
            this.J = 0;
          }
          for (var j = this.J; j < temp.length; j++) {
            var el = temp[j]
            if (typeof el === 'number') {
              this.J = j
              this._hasCallHasNext = true;
              return true;
            }
          }
        }
        return false
      }
    }