Question

如果输入了多个名称，我应该使用气泡排序按字母顺序列出它们。但是，我似乎无法使我的算法正确，因为我的输出以某种随机顺序给出。我的示例输入是

5 
Alice Hawking 
John Smith 
Stephen Hawking 
Alice Cooper 
Jean Smith

代码是

int main() {
    int number;
    char char1;
    char char2;
    int measure;
    int name1;
    int name2;
    int flag = 1;
    int count = 0, ter = 0;
    string tmp;

    cin >> number;

    string list[number+1][2];

    for (int i = 0; i < number; i++) {
        cin >> list[i][0] >> list[i][1];
    }

    // sorting first names
    for (int k = 0; k <= number; k++) {
        for (int i = 0; i < number - 1; i++) {
            // j cannot go beyond the length of the smallest first name
            // so measure will be the measure of the smallest first name
            if (list[i][0].size() < list[i+1][0].size()) {
                measure = list[i][0].size();
            } else {
                measure = list[i+1][0].size();
            }

            // convert the each letter of the string by converting 
            // string to char to int
            // flag is to indicate if the two comparisons are not the same
            // so that the if loop does not continue 
            for (int j = 0; flag && j < measure; j++) {
                char1 = list[i][0].at(j);
                char2 = list[i+1][0].at(j);
                name1 = (int)char1;
                name2 = (int)char2;
                if (name1 > name2) {
                    tmp = list[i][0];
                    list[i][0] = list[i+1][0];
                    list[i+1][0] = tmp;
                    flag = 0;
                }
            }
            flag = 1;
        }
    }

    // sorting last names
    for (int k = 0; k <= number; k++) {
        for (int i = 0; i < number - 1; i++) {
            // j cannot go beyond the length of the smallest last name
            // so measure will be the measure of the smallest last name
            if (list[i][1].size() < list[i+1][1].size()) {
                measure = list[i][0].size();
            } else {
                measure = list[i+1][0].size();
            }

            for (int j = 0; flag && j < measure; j++) {
                char1 = list[i][1].at(j);
                char2 = list[i+1][1].at(j);
                name1 = (int)char1;
                name2 = (int)char2;
                if (name1 > name2) {
                    tmp = list[i][1];
                    list[i][1] = list[i+1][1];
                    list[i+1][1] = tmp;
                    flag = 0;
                }
            }
            flag = 1;
        }
    }


    for (int t = 0; t < number; t++) {
        cout << list[t][0] << " " << list[t][1] << endl;
    }
}

Answer 1

以下是一些要注意的事项。

1比较字符串。

您是否有任何理由不仅仅使用string比较来比较名称？从// j cannot go到flag=的整个代码（交换除外）可以替换为

if (list[i][0] > list[i+1][0])

2比较两个循环中的第一个和最后一个。

检查同一循环中的姓氏和名字会好得多。类似的东西：

if (list[i][0] > list[i+1][0] || (list[i][0] == list[i+1][0] && list[i][1] > list[i+1][1]) {
}

注意：如果您想按姓氏和名字排序，请更改0和1。

3交换

目前，您正在独立于姓氏交换名字。这将导致你的名字混乱。同时交换它们。

按字母排序的气泡排序数组

1 个答案: