Question

我正在尝试比较两个数组(array1, array2)，如果array2中包含特定的键值，则array1中包含array2值的键值需要打印出来它的索引路径＆＃39;。

我几乎可以使用的代码，应用程序崩溃，因为在通过键时，array2超出范围，因为它包含的数组少于array1

如果其他数组较小，如何让代码查找匹配？

let array1 = [["aaa","bbb","ccc","ddd","eee"], ["fff","ggg","hhh","matched","iii"], ["lll","mmm","nnn","ooo","ppp"], ["666","777","888","999","000"] ] //4 elements

let countArray1 = array1.enumerate()

let array2 = [["111","222"], ["333","444"], ["matched","555"]] // 3 elements


for (index, element) in countArray1{

    let containedValue = array1[index].contains(array2[index][0])

    if (containedValue) == true{

        print("The index of the contained value is: ????") //error

    }
}

Answer 1

我会将较小的2D数组转换为1D数组，然后遍历较大数组的两个维度并使用NSArray.contains:

免责声明：此代码未经测试。 @Community：如果您偶然发现并发现错误或可以更好地编写，评论或编辑的内容。我比Swift更多地使用Obj-C。

//convert 2D array to 1D
func from2Dto1D(array: NSArray) -> NSArray {
    var newArr = NSMutableArray()
    for (d1) in array {
        for (d2) in d1 {
            newArr.append(d2)
        }
    }
    return newArr as NSArray
}
func hasMatch(array1: NSArray, array2: NSArray) -> boolean {
    var bigger: NSArray
    var smaller: NSArray
    if (array1.count > array2.count){
        bigger = array1
        smaller = from2Dto1D(array2)
    }else{
        bigger = array2
        smaller = from2Dto1D(array1)
    }
    for (d1) in bigger{
        for (item) in d1 {
            if (smaller.contains(item)){
                print("Match found");
                return true
            }
        }
    }
    return false
}

Answer 2

您可以展平两个数组，然后枚举它们以检查对象匹配的索引。

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Swift 2 - 检查数组是否包含来自另一个较小数组的值

2 个答案: