I am currently working on a problem that is making use of Levenshtein distance to calculate the optimal score between two strings.
I am successfully able to calculate the score for smaller strings. But when I try it with longer strings (exceeding length of more than 5000) i seem to be getting the wrong output. I am currently thinking if the issue has to do with memory allocation.
Below is my working
bool isVowel(char word){
return word == 'a' || word == 'e' || word == 'i' || word == 'o' || word == 'u';
}
int calculateMismatch(char first, char second){
int cost;
if((first != second) && (isVowel(first))){
cost = 3;
}else if((first != second) && (isVowel(first) && isVowel(second))){
cost = 1;
}else if((first != second) && (isVowel(second))){
cost = 3;
}else{
cost = 1;
}
return cost;
}
int getMinimum(int val1 , int val2, int val3){
if((val1 <= val2) && (val1 <= val3)){
return val1;
}
if((val2 <= val1) && (val2 <= val3)){
return val2;
}
if((val3 <= val1) && (val3 <= val2)){
return val3;
}
return 0;
}
int calculatingScore(string str1, string str2){
int sizestr1 = str1.size();
int sizestr2 = str2.size();
int cost;
vector<int> initial(sizestr2 + 1);
vector<vector<int> > result(sizestr1 + 1, initial);
result[0][0] = 0;
for(int i = 1; i < sizestr1 + 1; i++){
result[i][0] = i * 2;
}
for(int j = 1; j < sizestr2 + 1; j++){
result[0][j] = j * 2;
}
for(int j = 1; j <= sizestr2 ; j++){
for(int i = 1; i <= sizestr1 ; i++){
if(str1[i - 1] == str2[j - 1]){
cost = 0;
}else{
cost = calculateMismatch(str1[i - 1], str2[j - 1]);
}
int val1 = result[i - 1][j] + 2;
int val2 = result[i][j-1] + 2;
int val3 = result[i-1][j-1] + cost;
result[i][j] = getMinimum(val1, val2, val3);
}
}
int finalResult = result[sizestr1][sizestr2];
return finalResult;
}
答案 0 :(得分:2)
我不知道Levenshtein距离但你的calculateMismatch函数似乎错了。这应该是
int calculateMismatch(char first, char second){
int cost;
if((first != second) && (isVowel(first) && isVowel(second)))
{
cost = 3;
}else if((first != second) && (isVowel(first))){
cost = 1;
}else if((first != second) && (isVowel(second))){
cost = 3;
}else{
cost = 1;
}
return cost;
}