Question

我已经看过类似的问题而没有提供我正在寻找的答案，所以如果这被视为重复，我会提前道歉。我试图将数组{2，null，3}和{4,5,6}组合成{6,5,9}。对不起，如果问题很愚蠢。

Answer 1

你可以做这样的事情

public Integer[] arraySum(Integer[] array1, Integer[] array2) {
    if (array1.length != array2.length) {
        throw new IllegalArgumentException("Arrays should have the same size.");
    }
    Integer[] result = new Integer[array1.length];
    for (int i = 0; i < array1.length; i++) {
        result[i] = getValue(array1[i]) + getValue(array2[i]);
    }
    return result;
}

private int getValue(Integer integer) {
    return integer == null ? 0 : integer;
}

Answer 2

也许你主要需要：

a3[i] = (a1[i] == null ? a1[i] : 0) + (a2[i] == null ? a2[i] : 0);

Answer 3

func HitApi(callback: (NSDictionary) -> Void){
        let mapDict = [ "1":"First", "2":"Second"]
        let json = [ "title":"ABC" , "dict": mapDict ]
        let jsonData:NSData?

        do {
            jsonData = try NSJSONSerialization.dataWithJSONObject(json, options: .PrettyPrinted)
        }catch{
            jsonData = nil
        }


        // create post request
        let url = NSURL(string: "http://myserver.com")!
        let request = NSMutableURLRequest(URL: url)
        request.HTTPMethod = "POST"
        request.setValue("application/x-www-form-urlencoded; charset=utf-8", forHTTPHeaderField: "Content-Type")
        request.setValue("application/x-www-form-urlencoded; charset=utf-8", forHTTPHeaderField: "Accept")
        request.HTTPBody = jsonData

         var dict = ["output":""]

        let task = NSURLSession.sharedSession().dataTaskWithRequest(request){ data,response,error in
            if error != nil{
                dict["output"] = "An error"
                callback(dict)
            }
            do {

                let  data = try NSJSONSerialization.JSONObjectWithData(data!, options: NSJSONReadingOptions.AllowFragments) as? NSDictionary
                dict["output"] = NSString(data: data!, encoding: NSUTF8StringEncoding)
                callback(dict)

            }catch{
                dict["output"] = "error"
                callback(dict)
            }
        }

        task.resume()
}

Answer 4

数组是一个非常古老的概念，你应该看一下Array List

对于你的情况我知道我认为两个静态数组有3个元素

int[] A = {2, null, 3};
int[] B = {4, 5, 6} ;
int[] C= new  int[3]

for(int i=0;i<3;i++){
C[i] = (A[i]==null?A[i]:0) + (B[i]==null?B[i]:0);
}
//C array is your final array but make sure you remove null from your code

Answer 5

我认为你可能需要这个，试试这些代码行，他们可能会有所帮助

    int[] a = {2, null,3};
    int[] b = {4, 5,6};
    int c[a.length()];
    for(int i = 0;i < 3;i++){
       c[i] = (a[i]==null?a[i]:0) + (b[i]==null?b[i]:0)
    }

通过将它们的值逐个求和到一个数组中来合并两个整数数组

5 个答案: