One_Edit_Distance
2018-02-15 18:51
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题目描述:
Given two strings S and T, determine if they are both one edit distance apart.(判断两个字串能否只进行一次操作就使两字串相等。)
思路:(1)如果两个字串的长度之差大于1,那么就不可能只通过一次操作使两字串相等,故直接返回false。
(2)如果两个字串的长度之差等于1,那么两个字串在去掉长字串之中的一位之后两字串相等。
(3)如果两个字串的长度相等,那么两个字串之中也只能有一处不同。
public class One_Edit_Distance {
public static boolean isOneEditDistance(String s, String t)
{
int sLength = s.length();
int tLength = t.length();
int diff = Math.abs(sLength-tLength);
if(diff>1)
{
return false;
}
if(diff==1)
{
for(int i=0;i<Math.min(sLength, tLength);i++)
{
if(s.charAt(i)!=t.charAt(i))
return s.length()>t.length()?(s.substring(i+1).equals(t.substring(i))):(t.substring(i+1).equals(s.substring(i)));
}
//前面所有项均相同那么只剩多出来的那一项,将其删除则使两个字串相同。
return true;
}
if(diff==0)
{
int count = 0;
for(int i=0;i<sLength&&count<2;i++)
{
if(s.charAt(i)!=t.charAt(i))
count++;
}
return count<2;
}
return false;
}
public static void main(String[] args) {
String s = "ACE";
String t = "ABCE";
System.out.println(isOneEditDistance(s,t));
}
}
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