字符串操作：两个字符串的最长连续公共子串

public int findMatch(String number0, String number1){
if(number0 == null || number1 == null || !isNumeric(number0) || !isNumeric(number1)){
return 0;
}
if(number0.length() !=  number1.length()){
return -1;
}
int len = number0.length();
int[][] result = new int[len][len];
for(int i = 0; i < len; i++){
for(int j = 0; j < len; j++){
result[i][j] = 0;
}
}

int maxLen = 0;
int maxindex = 0;
int i = 0;
for(i=0;i<len;i++)
{
for(int j=0;j<len;j++)
{
if(number0.charAt(i) == number1.charAt(i))
{
if(i != 0 && j != 0)
{
result[i][j]=result[i-1][j-1]+1;
}
if(i==0||j==0)
{
result[i][j]=1;
}
if(result[i][j]>maxLen)
{
maxLen = result[i][j];
maxindex=i+1-maxLen;
}
}
}
}
return maxLen;
}