编辑距离
有两个字符串。第一个字符串是源字符串，第二个字符串是目标字符串。在此程序中，我们必须查找将第一个字符串转换为第二个字符串所需的编辑次数。
字符串的编辑可以是：插入一些元素，从第一个字符串中删除某些内容，或修改某些内容以转换为第二个字符串。
输入输出
Input: Two strings to compare. string 1: Programming string 2: Programs Output: Enter the initial string: Programming Enter the final string: Programs The number of changes required to convert Programming to Programs is 4
算法
editCount(initStr, finalStr, initLen, finalLen)
输入-初始和最终字符串及其长度。
输出-使initStr成为finalStr所需的编辑次数。
Begin
if initLen = 0, then
return finalLen
if finalLen := 0, then
return initLen
if initStr[initLen - 1] = finalStr[finalLen - 1], then
return editCount(initStr, finalStr, initLen – 1, finalLen - 1)
answer := 1 + min of (editCount(initStr, finalStr, initLen , finalLen - 1)),
(editCount(initStr, finalStr, initLen – 1, finalLen ),
(editCount(initStr, finalStr, initLen – 1, finalLen - 1)
return answer
End示例
#include<iostream>
using namespace std;
int min(int x, int y, int z) { //find smallest among three numbers
if(x < y) {
if(x < z)
return x;
else
return z;
}else {
if(y < z)
return y;
else
return z;
}
}
int editCount(string initStr , string finalStr, int initLen, intfinalLen) {
if (initLen == 0) //when initial string is empty, add all characters of final string
return finalLen;
if (finalLen == 0) //when final string is empty, delete all characters from initial string
return initLen;
//当最后一个字符匹配时，递归检查前一个字符
if (initStr[initLen-1] == finalStr[finalLen-1])
return editCount(initStr, finalStr, initLen-1, finalLen-1);
//如果找不到最后一个字符，则递归检查插入，删除和更新操作
return 1 + min ( editCount(initStr, finalStr, initLen, finalLen- 1), // insert
editCount(initStr, finalStr, initLen-1, finalLen), // delete
editCount(initStr, finalStr, initLen-1, finalLen-1) // update
);
}
int main() {
string initStr;
string finalStr;
cout << "Enter the initial string: "; cin >> initStr;
cout << "Enter the final string: "; cin >> finalStr;
cout << "The number of changes required to convert " << initStr << " to " << finalStr;
cout << " is " << editCount( initStr , finalStr, initStr.size(), finalStr.size()) << endl;
}输出结果
Enter the initial string: Programming Enter the final string: Programs The number of changes required to convert Programming to Programs is 4