目录
原题
小知识点:memset(a,'0',sizeof(a));
矩阵思想
最大公共子串
#include <stdio.h>
#include <string.h>
#define N 256
int f(const char* s1, const char* s2)
{
int a[N][N];
int len1 = strlen(s1);
int len2 = strlen(s2);
int i,j;
memset(a,0,sizeof(int)*N*N);
int max = 0;
for(i=1; i<=len1; i++){
for(j=1; j<=len2; j++){
if(s1[i-1]==s2[j-1]) {
a[i][j] = __________________________;
if(a[i][j] > max) max = a[i][j];
}
}
}
return max;
}
int main()
{
printf("%d\n", f("abcdkkk", "baabcdadabc"));
printf("%d\n", f("aaakkkabababa", "baabababcdadabc"));
printf("%d\n", f("abccbaacbcca", "ccccbbbbbaaaa"));
printf("%d\n", f("abcd", "xyz"));
printf("%d\n", f("ab", "ab"));
return 0;
}
二、
memset(a,'0',sizeof(a));
把a中的所有字符替换为‘0’
函数原型如下:
void *memset(void *s, int ch,int n);
函数解释:将s中前n个字节 用 ch 替换并返回 s将ch设置为0
三、矩阵思想是个啥(终于等到你
其实很简单,画个图就清楚了
原始状态,所有a[i][j]全是0
s2[0] | s2[1] | s2[2] | s2[3] | s2[4] | |
s1[0] | 0 | 0 | 0 | 0 | 0 |
s1[1] | 0 | 0 | 0 | 0 | 0 |
s1[2] | 0 | 0 | 0 | 0 | 0 |
s1[3] | 0 | 0 | 0 | 0 | 0 |
s1[4] | 0 | 0 | 0 | 0 | 0 |
发现有相等的字符,就让a[i][j]=1;
s2[0](x) | s2[1](x) | s2[2](a) | s2[3](b) | s2[4](c) | |
s1[0](e) | 0 | 0 | 0 | 0 | 0 |
s1[1](b) | 0 | 0 | 0 | 1 | 0 |
s1[2](a) | 0 | 0 | 1 | 0 | 0 |
s1[3](b) | 0 | 0 | 0 | 1 | 0 |
s1[4](c) | 0 | 0 | 0 | 0 | 1 |
最后计算出对角线的max,就是答案了!!
注意a[i][j]是int型,不是习惯上的char,注意他不是用来存放字符的!!