作为初学者，要学会举一反三，才能更好的掌握

今日看到一题squeeze: delete all characters stored in variable c from string s

题目很简单，删除s中的所有字符

函数接口定义：

void squeeze(char s[], int c);

裁判测试程序样例：

#include <stdio.h>

void squeeze(char s[], int c);

/* the length of a string */
int main(){
    char s[100];
    char c;
    int i;

    scanf("%s %c", s, &c);

    squeeze(s, c);
    for (i = 0; s[i] != '\0'; i++)
        putchar(s[i]);
    putchar('\n');

    return 0;
}
/* 请在这里填写答案 */

题解：

void squeeze(char s[], int c)
{
    int i, j;
	for (i = j = 0; s[i] != '\0'; ++i)
		if (s[i] != c)
			s[j++] = s[i];
	s[j] = '\0';
}

但如果是删除一个子串呢？

可以用BF算法或KMP算法解决

在这里我用BF算法删除一个子串，KMP算法找到子串位置

BF算法：

#include <string.h>
#include <stdio.h>

int squeeze(char s1[], char s2[]);
int main(){
    char s1[100];
    char s2[100];
    int i;
    int index;
    scanf("%s %s", s1, s2);
    index = squeeze(s1,s2);
    for (i = index;i<=strlen(s2) && s1[i+strlen(s2)]!='\0';i++)
    {
    	s1[i] = s1[i+strlen(s2)];
	}
	s1[i] = '\0';
    
    for (i = 0; s1[i] != '\0'; i++)
        putchar(s1[i]);
    putchar('\n');

    return 0;
} 
int squeeze(char s1[],char s2[])
{
    int i = 0,j = 0;
    while(i<strlen(s1) && j<strlen(s2))
    {
        if(s1[i] == s2[j]){
            ++i;
            ++j;
        }
        else{
            i = i-j+1;
            j = 0;//重新指向子串第一个
        }
    }
    if(j>=strlen(s2))
    {
        return i-strlen(s2);//返回子串位置
        
    }
    else return -1;
}

KMP算法：

#include <string.h>
#include <stdio.h>

int squeeze(char s1[], char s2[]);
void getnext(char s2[]);
int next[50] = {0};


int main(){
    char s1[100];
    char s2[100];
    int i;
    int index;
    scanf("%s %s", s1, s2);
    getnext(s2);
    index = squeeze(s1,s2);
    
    printf("%d",index);

    return 0;
}

int squeeze(char s1[],char s2[])
{
    int i = 0,j = 0;
    int len1,len2; 
	len1 = strlen(s1);
    len2 = strlen(s2);
    while(i<len1 &&j<len2)
    {
    	if(j==-1 || s1[i] == s2[j])
    	{
    		i++;
    		j++;
		}
		else j = next[j];
	}
	if(j>=len2)return i-len2;
	else return -1;
}

void getnext(char s2[])
{
	int i = 2,j = 0;
	next[0] = -1;
	next[1] = 0;
	while(i<strlen(s2))
	{
		if(j==-1 || s2[j]==s2[i-1])
		{
			next[i] = j+1;
			++i;
			++j;
			
		}
		else j = next[j];
	}
}