最長公共子串c語言-ag真人国际官网
❶ 求兩個輸入的字元串的最長公共子串
演算法:求兩個字元串的最長公共子串
原理:
(1) 將連個字元串分別以行列組成一個矩陣。
(2)。若該矩陣的節點對應的字元相同,則該節點值為1。
(3)當前字元相同節點的值 = 左上角(d[i-1, j-1])的值 1,這樣當前節點的值就是最大公用子串的長。
(s2)bcde
(s1)
a0000
b1000
c0200
d0030
3. 結果:只需以行號和最大值為條件即可截取最大子串
c# code:
[csharp]view plainprint?
publicstaticstringmylcs(strings1,strings2)
{
if(string.isnullorempty(s1)||string.isnullorempty(s2))
{
returnnull;
}
elseif(s1==s2)
{
returns1;
}
intlength=0;
intend=0;
int[,]a=newint[s1.length,s2.length];
for(inti=0;i
{
for(intj=0;j
{
intn=(i-1>=0&&j-1>=0)?a[i-1,j-1]:0;
a[i,j]=s1[i]==s2[j]?1 n:0;
if(a[i,j]>length)
{
length=a[i,j];
end=i;
}
}
}
returns1.substring(end-length 1,length);
}
❷ c璇璦 鏈闀垮叕鍏卞瓙涓
棣栧厛鎸囧嚭妤間富鐨勯敊璇
鏈闀跨殑鍏鍏卞瓙瀛楃︿覆 搴旇ユ敼鎴 鏈闀跨殑榪炵畫鍏鍏卞瓙瀛楃︿覆
涓嬮潰鏄絎﹀悎妤間富瑕佹眰鐨勫弬鑰冧唬鐮
//浣滆:hacker
//鏃墮棿:9.12.2006
#include
#include
void main()
{
char* x="aabcdababce";
char* y="12abcabcdace";
int m = strlen(x);
int n = strlen(y);
int i, j, k, l;
int maxlength = 0;
int start = 0;
int count = 0;//鐢ㄦ潵鍒ゆ柇鏄鍚﹀尮閰嶇殑鍙橀噺
for (i=1;i<=n;i )//鍖歸厤闀垮害鐨勫驚鐜
for (j=0;j
count = 0;
for (l=0;l if (y[j l] == x[k l])
count ;
if (count==i&&i>maxlength)
{
maxlength = i;//璁板綍鏈澶ч暱搴
start = j;//璁板綍鏈澶ч暱搴︾殑璧瘋搗浣嶇疆
}
}
//浣滆:hacker
//鏃墮棿:9.12.2006
#include
#include
void main()
{
char* x="aabcdababce";
char* y="12abcabcdace";
int m = strlen(x);
int n = strlen(y);
int i, j, k, l;
int maxlength = 0;
int start = 0;
int count = 0;//鐢ㄦ潵鍒ゆ柇鏄鍚﹀尮閰嶇殑鍙橀噺
for (i=1;i<=n;i )//鍖歸厤闀垮害鐨勫驚鐜
for (j=0;j
count = 0;
for (l=0;l if (y[j l] == x[k l])
count ;
if (count==i&&i>maxlength)
{
maxlength = i;//璁板綍鏈澶ч暱搴
start = j;//璁板綍鏈澶ч暱搴︾殑璧瘋搗浣嶇疆
}
}
if (maxlength==0)
printf("no answer");
else
for (i=0;i
}
}
涓嬮潰鏄鐪熸g殑鏈闀垮叕鍏卞瓙涓茬殑鍔ㄦ佽勫垝綆楁硶
//浣滆:hacker
//鏃墮棿:9.12.2006
#include
#include
int b[50][50];
int c[50][50];
void lcs(x,m,y,n)
char *x;
int m;
char *y;
int n;
{
int i;
int j;
for (i=1;i<=m;i ) c[i][0] = 0;
for (i=1;i<=n;i ) c[0][i] = 0;
c[0][0] = 0;
for (i=1;i<=m;i )
for (j=1;j<=n;j )
{
if (x[i-1] == y[j-1])
{
c[i][j] = c[i-1][j-1] 1;
b[i][j] = 1;
}
else
if (c[i-1][j] > c[i][j-1])
{
c[i][j] = c[i-1][j];
b[i][j] = 2;
}
else
{
c[i][j] = c[i][j-1];
b[i][j] = 3;
}
}
}
void show(i,j,x)
int i;
int j;
char* x;
{
if (i==0||j==0)
return;
if (b[i][j]==1)
{
show(i-1,j-1,x);
printf("%c",x[i-1]);
}
else
if (b[i][j]==2)
show(i-1,j,x);
else
show(i,j-1,x);
}
void main()
{
char* x="aabcdababce";
char* y="12abcabcdace";
int m = strlen(x);
int n = strlen(y);
lcs(x,m,y,n);
show(m,n,x);
}