CareerCup-BASE CASE AND BUILD_sbase case and-优快云博客

本文链接：https://blog.youkuaiyun.com/yellowxz/article/details/8841194

本文介绍了一种打印字符串所有可能排列的算法实现，并提供了一个C++示例程序。该算法通过递归方式逐步增加字符来生成所有可能的排列组合。

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这个题刷起来感觉比编程之美什么的easy多了

Description:

Solve the algorithm first for a base case (e g , just one element) Then, try to solve it for elements one and two, assuming that you have the answer for element one Then, try to solve it for elements one, two and three, assuming that you have the answer to elements one and two

Example:

Design an algorithm to print all permutations of a string For simplicity, assume all characters are unique.

Testing String: abcdefg

Case “a” --> {a}
Case “ab” --> {ab, ba}
Case “abc” --> ?

#include <iostream>
using namespace std;

void print(char* prefix, char* data, int index, int length)
{
     if(index >= length)
     {
         static int count = 0;
         printf("%d: %s\n", count++, prefix);
         return;
     }
     int prelength = strlen(prefix);
     for(int i=0; i<prelength+1; i++)
     {
         char* s = new char[prelength + 2];;
         int count = 0, j;
         for(j=0; j<prelength+1; j++)
         {
             if(j == i)
                 s[j] = data[index];
             else
                 s[j] = prefix[count++];
         }
         s[j] = '\0';
         print(s, data, index+1, length);
     }
};

int main()
{
    char* data = "abcd";
    print("", data, 0, strlen(data));
    system("pause"); 
};