幾個經典算法研究

最近研究幾個經典算法，比如二分查找，大數相加等。加強一下編碼思路訓練，這幾個算法都不是很難，基本可以在演算中得到代碼實現步驟，如下：

#include<stdio.h>
#include<string>
#include<vector>
#include <iostream>
using namespace std;
#define MAX 1010
const int NotFound = -1;


int Bsearch(const vector<string> &v, int start, int stop, string key) //二分查找，遞歸實現。當然之間用循環就可以了！
{
    if (start > stop)
    {
        return NotFound;
    }
    int mid = (start + stop)/2;
    if (key == v[mid])
    {
        return mid;
    }
    else if (key < v[mid])
    {
        return Bsearch(v, start, mid - 1, key);
    }
    else
    {
        return Bsearch(v, mid + 1, stop, key);
    }
}
string bignumadd(const string str1, const string str2)  //大數相加
{
    int a[MAX] = {0}, b[MAX] = {0};
    int len1, len2, i, k, up = 0;
    int len = 0;
    char buf[512] = {0};
    len1 = str1.length();
    len2 = str2.length();
    for(i = len1 - 1, k = 0; i >= 0; --i)
        a[k++] = str1[i] - '0';
    for(i = len2 - 1, k = 0; i >= 0; --i)
        b[k++] = str2[i] - '0';
    for(i = 0, up = 0; i < MAX; ++i)
    {
        a[i] = a[i] + b[i] + up;
        up = a[i] / 10;
        a[i] %= 10;
    }
    for(i = MAX -1; i >= 0; --i)
        if(a[i]) break;
    len += sprintf(buf + len, "%s + %s = ", str1.c_str(), str2.c_str());
    for(k = i; k >= 0; --k)
    {
        len += sprintf(buf + len ,"%d", a[k]);
    }
    string str(buf);
    printf("%s\n", buf);
    return str;
}
void RecPermute(string sofar, string rest)  //全排列（遞歸算法）
{
    if (rest == "")
    {
       cout << sofar << endl;
    }
    else
    {
        for (int i = 0; i < rest.length(); i++)
        {
            string next = sofar + rest.at(i);
            string remaining = rest.substr(0, i)
                             + rest.substr(i + 1, rest.length() - i);
            RecPermute(next, remaining);
        }
    }
}
void move(int n,char src,char dest,char tmp) //漢諾塔問題
{
    if(n==1)
        printf("\t%c->%c\n",src,tmp);    //當n只有1個的時候直接從a移動到c
    else
    {
        move(n-1,src,tmp,dest);            //第n-1個要從a通過c移動到b
        printf("\t%c->%c\n",src,tmp);
        move(n-1,dest,src,tmp);            //n-1個移動過來之後b變開始盤，b通過a移動到c，這邊很難理解
    }
}
int main()
{
   RecPermute("","ABCD");

    return 0;
}

******************************************************************************************************************

qsort排序函數在結構體數組中的運用

比如實現如下要求：

先按照學生姓名字典序排序，學生姓名一樣的，按照分數從大到小排序。實例代碼如下：

//============================================================================
// Name        : CTest.cpp
// Author      : @CodingGeek
// Version     : 1.0
// Copyright   : Your copyright notice
// Description : 幾種常見的STL容器學習研究
//============================================================================
#include <iostream>
#include <string.h>
#include <stdlib.h>
using namespace std;

typedef struct
{
  int id;
  char name[20];
  int score;
}T_Member;

T_Member g_MemIdTbl[] =
{
 {1001, "zhangsan", 79},
 {1004, "lisi", 78},
 {1008, "wangwu", 85},
 {1006, "zhaoliu", 87},
 {1007, "sunqian", 67},
 {1005, "wangyong", 78},
 {1009, "caoxie", 76},
 {1003, "wangwu", 90},
};

#define SIZE  sizeof(g_MemIdTbl) / sizeof(g_MemIdTbl[0])

int compare_fun(const void *a, const void *b)
{
    T_Member *pa = (T_Member*)a;
    T_Member *pb = (T_Member*)b;

    if (strcmp(pa->name,pb->name) != 0)
    {
         return strcmp(pa->name,pb->name);
    }
    else
    {
        return pb->score - pa->score;
    }
}

int main()
{

    qsort(g_MemIdTbl, SIZE, sizeof(g_MemIdTbl[0]),compare_fun);

    for (unsigned i = 0; i < SIZE; i++)
    {
      cout << g_MemIdTbl[i].id<<":"<<g_MemIdTbl[i].name <<":"
           << g_MemIdTbl[i].score << endl;
    }
    return 0;
}