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博客围绕面向过程、字符串和指针展开。面向过程部分介绍 Lab3 - 03 去重排序的三种写法及时间复杂度；字符串部分提及 Lab04_02 小写变大写排序；指针部分涉及矩阵转置。

面向过程

Lab3-03 去重排序
写法1，时间复杂度 $O(n^2)$

#include<stdio.h>
#include<iostream>
#include<vector>
using namespace std;

int main()
{
    char order;
    cin >> order;
    vector<int> vi;
    char ch = ' ';
    int x = 0;
    while(ch!='\n'){
        cin >> x;
        vi.push_back(x);
        ch = cin.get();
    }

    //先排序
    if(order == 'A') {  //升序
        for (int i = 0; i < vi.size(); i++) {
            int most = 1000;
            int most_i = 0;
            for (int j = i; j < vi.size(); j++) { //每次找到最小的放到前面
                if (vi[j] < most) {
                    most = vi[j];
                    most_i = j;
                }
            }
            int temp = vi[i];  //交换
            vi[i] = vi[most_i];
            vi[most_i] = temp;
        }
    }

    else if(order == 'D') {  //降序
        for (int i = 0; i < vi.size(); i++) {
            int most = 0;
            int most_i = 0;
            for (int j = i; j < vi.size(); j++) { //每次找到最小的放到前面
                if (vi[j] > most) {
                    most = vi[j];
                    most_i = j;
                }
            }
            int temp = vi[i];  //交换
            vi[i] = vi[most_i];
            vi[most_i] = temp;
        }
    }

    //再去重
    for(int i=0; i<vi.size(); i++){
        int k = 1;
        while(vi[i]==vi[i+k]&&(i+k)<vi.size()) k++;
        if(k>1){
            for(int j=i+1;j<vi.size()-k+1;j++) vi[j] = vi[j+k-1];
            for(int p=0;p<k-1;p++) vi.pop_back();
        }
    }

    for(int i=0;i<vi.size()-1; i++) cout << vi[i] << " ";
    cout << vi[vi.size()-1];

}

写法2，哈希，时间复杂度 $O (M)$

#include<stdio.h>
#include<iostream>
#include<vector>
using namespace std;

int main()
{
    char order;
    cin >> order;
    vector<int> vi;
    char ch = ' ';
    int x = 0;
    while(ch!='\n'){
        cin >> x;
        vi.push_back(x);
        ch = cin.get();
    }

    int a[1001];
    for(int i=0; i<1001; i++) a[i] = 0;
    for(int i=0; i<vi.size(); i++) a[vi[i]] = 1;

    int num=0;
    if(order=='A'){
        for(int i=1;i<=1000;i++){
            if(a[i]==1){
                vi[num] = i;
                num++;
            }
        }
    }
    else if(order=='D'){
        for(int i=1000; i>0; i--){
            if(a[i]==1){
                vi[num] = i;
                num++;
            }
        }
    }

    for(int i=0;i<num-1;i++) cout<<vi[i]<<" ";
    cout << vi[num-1];
}

写法3，用到set

#include<stdio.h>
#include<iostream>
#include<vector>
#include<set>
using namespace std;

int main()
{
    char order;
    cin >> order;
    vector<int> vi;
    char ch = ' ';
    int x = 0;
    while(ch!='\n'){
        cin >> x;
        vi.push_back(x);
        ch = cin.get();
    }

    set<int> st;
    for(int i=0;i<vi.size();i++){
        st.insert(vi[i]);
    }

    if(order=='A'){
        set<int>::iterator it=st.begin();
        cout << *it;
        for(it++;it!=st.end();it++){
            cout << ' ' << *it;
        }
    }
}

字符串

Lab04_02 小写变大写

#include<stdio.h>
#include<iostream>
using namespace std;

int main()
{
    string s;
    cin >> s;
    for(int i=0; i<s.size(); i++){
        if(s[i]<='z'&&s[i]>='a') s[i]=s[i]-32;
    }
    
    cout << s;
}

排序

#include<iostream>
using namespace std;

void swp(string& s, int i, int j){
    char swp;
    swp = s[i];
    s[i] = s[j];
    s[j] = swp;
}

int find_last(string s, int end){ //找到end前最后一个字母的位置
    for (int i = end-1; i >= 0; i--)
    {
        if (s[i] >= 'A'&&s[i] <= 'Z')
            return i;
    }
    return -1;
}

int main(){
    string s;
    cout << "输入字符串" << endl;
    cin >> s;

    //先找到字符串中第一个出现大写字母与最后出现大写字母的位置
    int begin=0;
    int end=0;
    for(int i=0; i<s.size(); i++){
        if(s[i]>='A'&&s[i]<='Z'){
            begin = i;
            break;
        }
    }
    for(int j=0; j<s.size(); j++){
        if(s[j]>='A'&&s[j]<='Z') {
            end = j;
        }
    }

    //从begin到end，每次选出最大的字符串放到后面
    while(end>begin){
        char max = s[begin];
        int max_i = begin;
        for(int i=begin; i<=end; i++){
            if(s[i]>max&&s[i]>='A'&&s[i]<='Z'){
                max=s[i];
                max_i = i;
            }
        }
        swp(s, max_i, end);
        end = find_last(s, end);
    }

    cout << "输出已排序字符串" << endl;
    cout << s << endl;
    return 0;
}

指针

矩阵转置

#include<iostream>
#include<string>
using namespace std;

void trans_mtx(int **a, int n)
{
    int b[n][n];
    for(int i = 0; i < n; i++){
        for(int j=0; j<n; j++){
        b[i][j] = a[i][j];
        }
    }

    for(int i = 0; i < n; i++){
        for(int j=0; j<n; j++){
        a[i][j] = b[j][i];
        }
    }
}

void print_mtx( int **a, int n)
{
 int i;
 for(i = 0; i < n; i++){
    for(int j=0; j<n; j++){
        cout << a[i][j] << "";
    }
    cout << endl;
 }
}

int main()
{
    char* p[3];
    char str[][5] = {"abc","def","ghi"};

    int* addre[3];
    int a[3][3] = {{1,2,3},
                    {1,2,3},
                    {1,2,3}};

    addre[0] = &a[0][0];
    addre[1] = a[1];
    addre[2] = a[2];
    trans_mtx(addre, 3);
    print_mtx(addre, 3);
    return 0;

}