Question

数组的类型为String 。由于添加＆＃34;＆＃34;需要耗费时间，因此我将其编写为Int。遗憾。

我有两个阵列说var array1 = [[1,2,3,4,5,6,7,8,9]]和

var array2 = [[1,2,3,4],
              [2,3,4,5],
              [2,4,5,6],
              [1,2,3,4,5,6,7,8,9],
              [1,2,3,4,5,6,7,8],
              [2,3,4,5,6,7,8]]

我必须将array2的每个数组元素与array1进行比较，并插入＆＃39; - ＆＃39;元素不匹配的地方。像这样，

var array2 = [[1,2,3,4,-,-,-,-,-],
              [-,2,3,4,5,-,-,-,-],
              [-,2,-,4,5,6,-,-,-],
              [1,2,3,4,5,6,7,8,9],
              [1,2,3,4,5,6,7,8,-],
              [-,2,3,4,5,6,7,8,-]]

我试图遍历array2中的每个数组并将其与array1进行比较，比较索引并插入＆＃39; - ＆＃39;到索引位置我，但我得到了意想不到的结果。

更新

 for item in array2{
var elementsArray = item
for i  in stride(from: 0, to: elementsArray.count, by: 1) {
    if elementsArray[i] != array1[i]
    {
        elementsArray.insert("-", at: i)
    }
    print("elemnt array.....", elementsArray, "\n\n")
}
}

我曾想过将array1的每个数组与array1进行count比较，找到不常见元素的索引，然后插入＆＃39; - ＆＃39;在那个指数位置。这种方法对吗？请帮帮我。

Answer 1

您需要一个新数组，其中array2的每一行都被替换为array1 let array1 = [1,2,3,4,5,6,7,8,9] let array2 = [[1,2,3,4], [2,3,4,5], [2,4,5,6], [1,2,3,4,5,6,7,8,9], [1,2,3,4,5,6,7,8], [2,3,4,5,6,7,8]] let filled = array2.map { row in array1.map { row.contains($0) ? String($0) : "-" } } for row in filled { print(row) }，行中最初不存在的元素替换为“ - ”：

Set(row)

输出：

["1", "2", "3", "4", "-", "-", "-", "-", "-"]
["-", "2", "3", "4", "5", "-", "-", "-", "-"]
["-", "2", "-", "4", "5", "6", "-", "-", "-"]
["1", "2", "3", "4", "5", "6", "7", "8", "9"]
["1", "2", "3", "4", "5", "6", "7", "8", "-"]
["-", "2", "3", "4", "5", "6", "7", "8", "-"]

对于大型数组，可以通过创建elementsArray来改进更快的遏制检查，或利用这些元素正在递增。

您的方法无效，因为static 迭代时修改它。

比较两个数组并找到SWIFT 3中不常见元素的索引

1 个答案: