let arr1 = [["aaa","111"],["bbb","222"],["ccc","333"]]
let arr2 = [["ddd","444"],["eee","555"],["fff","666"]]
let arr3 = arr1 + arr2
print(arr3) // [["aaa", "111"], ["bbb", "222"], ["ccc", "333"], ["ddd", "444"], ["eee", "555"], ["fff", "666"]]
在您的情况下,具有特定要求。
let arr4 = zip(arr1, arr2).reduce([]) { (var arr, p:(Array<String>, Array<String>)) -> [[String]] in
arr.append(p.0)
arr.append(p.1)
return arr
}
print(arr4) // [["aaa", "111"], ["ddd", "444"], ["bbb", "222"], ["eee", "555"], ["ccc", "333"], ["fff", "666"]]
根据您在笔记中提到的需求,您可以按照以下方式添加其余的值(请先将let arr4更改为var arr4!!!)
var i = arr4.count / 2
while i < arr1.count {
arr4.append(arr1[i++])
}
while i < arr2.count {
arr4.append(arr2[i++])
}
print(arr4)
这将为您提供
[["aaa", "111"], ["ddd", "444"], ["bbb", "222"], ["eee", "555"], ["ccc", "333"], ["fff", "666"], ["zzz", "755"]]
即使其中一个数组为空,它也应该能够正常工作。
let arr1 = [["aaa","111"],["bbb","222"],["ccc","333"],["zzz","755"]]
-let arr2 = [["ddd","444"],["eee","555"],["fff","666"]]
- SNos