（排序法之快速法）对N行N列二维数组的每一行排序，偶数行(0当作偶数)由小到大排序，奇数行由大到小排序

对N行N列二维数组的每一行排序，偶数行(0当作偶数)由小到大排序，奇数行由大到小排序

////main.c#include "stdafx.h"#include "head.h"int main(){    /**********定义数组并原样输出*************/    char a[4][7] = { 'g','f','e','d','c','b','a',                     '7','6','5','4','3','2','1',                     '1','2','3','4','5','6','7',                     'q','w','e','r','t','y','u', };    for (int i = 0; i < 4; i++)    {        printf("\n第%d行为:", i + 1);        for (int j = 0; j < 7; j++)//输出            printf("%c ", *(a[i] + j));     }    puts("");//换行    /******开始处理数组并输出*********/    for (int i = 0; i < 4; i++)    {        if (i /2 == 0)        {            printf("\n第%d行从小到大排序为:", i+1);            quick_rank_string_small_to_large(a[i], 0, 6);//排序，head.c文件中定义        }        else        {            printf("\n第%d行从大到小排序为:", i+1);            quick_rank_string_large_to_small(a[i], 0, 6);//排序        }        for (int j = 0; j < 7; j++)//输出            printf("%c ", *(a[i] + j));    }}//************************我是分隔符*******************************//***************************************************************//head.c#include "stdafx.h"#include "head.h"//快速法//字符串从小到大排序,left为要排序的字符串的首字母下标，right为末字母下标void quick_rank_string_small_to_large(char *a, int left, int right){    int mid = *(a + (left + right) / 2);    int left_go = left, right_go = right;    do    {        while (a[left_go] < mid&&left_go < right)            left_go++;//舍去        while (a[right_go] > mid&&right_go > left)            right_go--;        if (left_go <= right_go)        {            int tmp = *(a + left_go);            *(a + left_go) = *(a + right_go);            *(a + right_go) = tmp;            left_go++; right_go--;        }    } while (left_go <= right_go);    if (left_go < right)        quick_rank_string_small_to_large(a, left_go, right);    if (right_go > left)        quick_rank_string_small_to_large(a, left, right_go);}//快速法//字符串从大到小排序,left为要排序的字符串的首字母下标，right为末字母下标void quick_rank_string_large_to_small(char *a, int left, int right){    int mid = *(a + (left + right) / 2);    int left_go = left, right_go = right;    do    {        while (a[left_go] > mid&&left_go < right)            left_go++;//舍去        while (a[right_go] < mid&&right_go > left)            right_go--;        if (left_go <= right_go)//满足条件交换        {            int tmp = *(a + left_go);            *(a + left_go) = *(a + right_go);            *(a + right_go) = tmp;            left_go++; right_go--;        }    } while (left_go <= right_go);    if (left_go < right)        quick_rank_string_large_to_small(a, left_go, right);//迭代    if (right_go > left)        quick_rank_string_large_to_small(a, left, right_go);//迭代}//************************我是分隔符*******************************//***************************************************************#pragma oncevoid quick_rank_string_small_to_large(char *a, int left, int right);//快速法，从小到大排序void quick_rank_string_large_to_small(char *a, int left, int right);//快速法，从大到小排序

结果如图：