一个含有字符串的数组，例如:["a","b","a","c","b","d","a"],需要得到一个新数组 ["a1","b1","a2","c","b2","d","a3"] c#

你可以通过遍历原数组，并使用 Dictionary 来统计每个字符串出现的次数，然后再次遍历原数组，根据每个字符串出现的次数，构造新的字符串数组。以下是一个示例代码：

string[] originalArray = new string[] { "a", "b", "a", "c", "b", "d", "a" };
Dictionary<string, int> countDict = new Dictionary<string, int>();
string[] newArray = new string[originalArray.Length];

// 统计每个字符串出现的次数
for (int i = 0; i < originalArray.Length; i++)
{
    string str = originalArray[i];
    if (countDict.ContainsKey(str))
    {
        countDict[str]++;
    }
    else
    {
        countDict[str] = 1;
    }

    int count = countDict[str];
    newArray[i] = str + count;
}

// 输出新的字符串数组
foreach (string str in newArray)
{
    Console.Write(str + " ");
}

相关问题

c#中 一个含有字符串的数组，例如:["a","b","a","c","b","d","a"],需要得到一个新数组 ["a1","b1","a2","c","b2","d","a3"]

你可以使用以下代码来实现你的需求：

string[] array = { "a", "b", "a", "c", "b", "d", "a" };
List<string> newArray = new List<string>();

Dictionary<string, int> countDict = new Dictionary<string, int>();

foreach (string item in array)
{
    if (countDict.ContainsKey(item))
    {
        countDict[item]++;
        newArray.Add(item + countDict[item]);
    }
    else
    {
        countDict[item] = 1;
        newArray.Add(item + countDict[item]);
    }
}

string[] resultArray = newArray.ToArray();

这段代码首先创建了一个空的`List<string>`来存储新的数组。然后，使用一个`Dictionary<string, int>`来记录每个字符串出现的次数。接下来，遍历原始数组中的每个元素，如果该元素已经在字典中存在，则将其对应的计数加1，并将带有计数的字符串添加到新数组中；如果该元素在字典中不存在，则将其添加到字典中，并将带有计数的字符串添加到新数组中。最后，将新数组转换为字符串数组并返回。

个含有字符串的数组，例如:["a","b"."a” ，b”，d”.a”1,1].需要得到一个新数组 ["a1""b1""a2""c","b2""d""a3""11""12"]

我认为输入的数组中有一些错误，应该是这样的：["a","b","a","b","d","a","1","1"]，而不是["a","b"."a” ，b”，d”.a”1,1]。在这个假设下，以下是一个可以得到新数组的实现：

def get_new_array(arr):
    d = {}
    new_arr = []
    for s in arr:
        if s in d:
            d[s] += 1
            new_arr.append(s + str(d[s]))
        else:
            d[s] = 1
            new_arr.append(s + str(d[s]))
    return new_arr

arr = ["a","b","a","b","d","a","1","1"]
new_arr = get_new_array(arr)
print(new_arr) # ["a1","b1","a2","b2","d1","a3","11","12"]

这个实现中，我们用一个字典 `d` 来记录每个字符串出现的次数，然后遍历输入数组，对于每个字符串，如果它已经出现过，就在它后面加上它出现的次数，否则就把它加入字典，并在后面加上 1。最后返回新数组。