Rotate/Shift Dartlang 中的列表?
Rotate/Shift a list in Dartlang?
Dart 中是否有 better/faster 旋转列表的方法?
List<Object> rotate(List<Object> l, int i) {
i = i % l.length;
List<Object> x = l.sublist(i);
x.addAll(l.sublist(0, i));
return x;
}
可以简化一点
List<Object> rotate(List<Object> list, int v) {
if(list == null || list.isEmpty) return list;
var i = v % list.length;
return list.sublist(i)..addAll(list.sublist(0, i));
}
如果您想 shift
而不是 rotate
,您可以简单地使用 removeAt
函数:
List<int> list = [ 1, 2, 3 ];
int firstElement = list.removeAt(0);
print(list); // [ 2, 3 ]
print(firstElement); // 1
来自文档:
Removes the object at position [index] from this list.
This method reduces the length of this by one and moves all later objects down by one position.
Returns the removed value.
The [index] must be in the range 0 ≤ index < length. The list must be growable.
Here 是一些更有用的 JS 垫片。
Dart 中是否有 better/faster 旋转列表的方法?
List<Object> rotate(List<Object> l, int i) {
i = i % l.length;
List<Object> x = l.sublist(i);
x.addAll(l.sublist(0, i));
return x;
}
可以简化一点
List<Object> rotate(List<Object> list, int v) {
if(list == null || list.isEmpty) return list;
var i = v % list.length;
return list.sublist(i)..addAll(list.sublist(0, i));
}
如果您想 shift
而不是 rotate
,您可以简单地使用 removeAt
函数:
List<int> list = [ 1, 2, 3 ];
int firstElement = list.removeAt(0);
print(list); // [ 2, 3 ]
print(firstElement); // 1
来自文档:
Removes the object at position [index] from this list.
This method reduces the length of this by one and moves all later objects down by one position.
Returns the removed value.
The [index] must be in the range 0 ≤ index < length. The list must be growable.
Here 是一些更有用的 JS 垫片。