您可以通过添加分隔符项这样展开。

list.expand((element)=>[...element,0])

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更新

list.reduce((a,b)=>[...a,0,...b])

型

一种解决方案是对Iterable进行扩展，如expandWithSeparator，并支持分隔符值：

void main() {
  List<List<int>> list = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
  ];

  print(list.expandWithSeparator((element) => element, 0).toList());
  // [1, 2, 3, 0, 4, 5, 6, 0, 7, 8, 9]
}

extension<E> on Iterable<E> {
  Iterable<T> expandWithSeparator<T>(
    Iterable<T> Function(E element) toElements,
    T separator,
  ) sync* {
    bool first = true;

    for (final element in this) {
      if (first) {
        first = false;
      } else {
        yield separator;
      }
      yield* toElements(element);
    }
  }
}

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List<List<int>> list = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
  ];
list.fold([], (total, item) => [...total, if (total.isNotEmpty) 0, ...item]);

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