Ios 将int（[3,6]）的数组转换为[[1,2,3]，[1,2,3,4,5,6]]

例如：-整数数组-[3,3,6]到[[1,2,3]，[1,2,3]，[1,2,3,4,5,6]]

lazy var arrCount = 3
lazy var numberCount = [Int]()
lazy var someArr = [SomeArr]()

for i in 0 ..< self!. arrCount.count {
  self?.numberCount.append(self!.arrCount[I].someArr!.count)
}
print(self?.numberCount.count) // now response based on above example prints [3,3,6]

lazy var arrCount=3
惰性变量numberCount=[Int]（）
lazy var someArr=[someArr]（）
因为我在0..

响应[3,3,6]-[1,2,3]，[1,2,3]，[1,2,3,4,5,6]]

lazy var arrCount = 3
lazy var numberCount = [Int]()
lazy var someArr = [SomeArr]()

for i in 0 ..< self!. arrCount.count {
  self?.numberCount.append(self!.arrCount[I].someArr!.count)
}
print(self?.numberCount.count) // now response based on above example prints [3,3,6]

---

提前感谢。

您可以使用

map

方法

var arr = [3, 6];
print(arr.map { Array(1...$0) })

输出

[[1, 2, 3], [1, 2, 3, 4, 5, 6]]

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@SaiKishore，你指的是总数9？那么只需使用一种方法，通过索引访问数组项即可。