LQ0219 三部排序程序填空
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题目来源:蓝桥杯2013初赛 C++ B组F题
题目描述
本题为代码补全填空题,请将题目中给出的源代码补全,并复制到右侧代码框中,选择对应的编译语言(C/Java)后进行提交。若题目中给出的源代码语言不唯一,则只需选择其一进行补全提交即可。复制后需将源代码中填空部分的下划线删掉,填上你的答案。提交后若未能通过,除考虑填空部分出错外,还需注意是否因在复制后有改动非填空部分产生错误。
一般的排序有许多经典算法,如快速排序、希尔排序等。
但实际应用时,经常会或多或少有一些特殊的要求。我们没必要套用那些经典算法,可以根据实际情况建立更好的解法。
比如,对一个整型数组中的数字进行分类排序:
使得负数都靠左端,正数都靠右端,0 在中部。注意问题的特点是:负数区域和正数区域内并不要求有序。可以利用这个特点通过 1 次线性扫描就结束战斗!!
以下的程序实现了该目标。
其中 x 指向待排序的整型数组,len 是数组的长度。
请分析代码逻辑,并推测划线处的代码。
源代码
C
#include <stdio.h>
void show(int* x, int len)
int i;
for(i=0; i<len; i++)
printf("%d,",x[i]);
printf("\\n");
void sort3p(int* x, int len)
int p = 0;
int left = 0;
int right = len-1;
while(p<=right)
if(x[p]<0)
int t = x[left];
x[left] = x[p];
x[p] = t;
left++;
p++;
else if(x[p]>0)
int t = x[right];
x[right] = x[p];
x[p] = t;
right--;
//p++;
else
_______________;
show(x, len);
int main()
int a[] = -1,0,1,-2,0,2,-3,0,0,3,-4,-5,4,-6,0,5,6;
int b[] = -1,0,-1,-2,0,-2,-3,0,0,-3,-4,-5,-4,-6,0,-5,-6;
int c[] = 1,0,1,2,0,2,3,0,0,3,4,5,4,6,0,5,6;
sort3p(a,sizeof(a)/sizeof(int));
sort3p(b,sizeof(b)/sizeof(int));
sort3p(c,sizeof(c)/sizeof(int));
return 0;
Java
import java.util.*;
public class Main
static void sort(int[] x)
int p = 0;
int left = 0;
int right = x.length-1;
while(p<=right)
if(x[p]<0)
int t = x[left];
x[left] = x[p];
x[p] = t;
left++;
p++;
else if(x[p]>0)
int t = x[right];
x[right] = x[p];
x[p] = t;
right--;
//p++;
else
______________;
show(x);
static void show(int[] x)
for(int i=0; i<x.length; i++)
System.out.print(x[i] + ",");
System.out.println();
public static void main(String[] args)
//int[] x = 25,18,-2,0,16,-5,33,21,0,19,-16,25,-3,0;
sort(new int[]-1,0,1,-2,0,2,-3,0,0,3,-4,-5,4,-6,0,5,6);
sort(new int[]-1,0,-1,-2,0,-2,-3,0,0,-3,-4,-5,-4,-6,0,-5,-6);
sort(new int[]1,0,1,2,0,2,3,0,0,3,4,5,4,6,0,5,6);
问题分析
C语言程序填入“p++”
Java语言程序也是填入“p++”
AC的C语言程序如下:
#include <stdio.h>
void show(int* x, int len)
int i;
for(i=0; i<len; i++)
printf("%d,",x[i]);
printf("\\n");
void sort3p(int* x, int len)
int p = 0;
int left = 0;
int right = len-1;
while(p<=right)
if(x[p]<0)
int t = x[left];
x[left] = x[p];
x[p] = t;
left++;
p++;
else if(x[p]>0)
int t = x[right];
x[right] = x[p];
x[p] = t;
right--;
//p++;
else
p++;
show(x, len);
int main()
int a[] = -1,0,1,-2,0,2,-3,0,0,3,-4,-5,4,-6,0,5,6;
int b[] = -1,0,-1,-2,0,-2,-3,0,0,-3,-4,-5,-4,-6,0,-5,-6;
int c[] = 1,0,1,2,0,2,3,0,0,3,4,5,4,6,0,5,6;
sort3p(a,sizeof(a)/sizeof(int));
sort3p(b,sizeof(b)/sizeof(int));
sort3p(c,sizeof(c)/sizeof(int));
return 0;
AC的Java语言程序如下:
import java.util.*;
public class Main
static void sort(int[] x)
int p = 0;
int left = 0;
int right = x.length-1;
while(p<=right)
if(x[p]<0)
int t = x[left];
x[left] = x[p];
x[p] = t;
left++;
p++;
else if(x[p]>0)
int t = x[right];
x[right] = x[p];
x[p] = t;
right--;
//p++;
else
p++;
show(x);
static void show(int[] x)
for(int i=0; i<x.length; i++)
System.out.print(x[i] + ",");
System.out.println();
public static void main(String[] args)
//int[] x = 25,18,-2,0,16,-5,33,21,0,19,-16,25,-3,0;
sort(new int[]-1,0,1,-2,0,2,-3,0,0,3,-4,-5,4,-6,0,5,6);
sort(new int[]-1,0,-1,-2,0,-2,-3,0,0,-3,-4,-5,-4,-6,0,-5,-6);
sort(new int[]1,0,1,2,0,2,3,0,0,3,4,5,4,6,0,5,6);
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