#leetcode#One Edit Distance
2015-07-20 16:18
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Given two strings S and T, determine if they are both one edit distance apart.
分析:one edit distance,要么S,T长度相同, 只有一对不同char,那么只要跳过这对不同的char,应该能匹配完两个String的;
要么S,T长度相差一,其他部分都匹配,则只要跳过长度长的那个String的不匹配字符继续匹配,如果可以匹配完两个String,则返回true
时间复杂度: 只需一遍扫描, O(n)
又观察了一下,其实shift == 1 的情况下没有必要单独讨论 if(i == s.length()),可以整合到下面的判断中, 改进如下
public class Solution {
public boolean isOneEditDistance(String s, String t) {
// make s to be the shorter String
if(s.length() > t.length()){
return isOneEditDistance(t, s);
}
if(t.length() - s.length() > 1){
return false;
}
int shift = t.length() - s.length();
int i = 0;
while(i < s.length() && s.charAt(i) == t.charAt(i)){
i++;
}
if(shift == 0){
i++;
while(i < s.length() && s.charAt(i) == t.charAt(i)){
i++;
}
return i == s.length();
}else{
while(i < s.length() && s.charAt(i) == t.charAt(i + 1)){
i++;
}
return i == s.length();
}
}
}
分析:one edit distance,要么S,T长度相同, 只有一对不同char,那么只要跳过这对不同的char,应该能匹配完两个String的;
要么S,T长度相差一,其他部分都匹配,则只要跳过长度长的那个String的不匹配字符继续匹配,如果可以匹配完两个String,则返回true
时间复杂度: 只需一遍扫描, O(n)
public class Solution { public boolean isOneEditDistance(String s, String t) { // make s to be the shorter String if(s.length() > t.length()){ return isOneEditDistance(t, s); } if(t.length() - s.length() > 1){ return false; } int shift = t.length() - s.length(); int i = 0; while(i < s.length() && s.charAt(i) == t.charAt(i)){ i++; } if(shift == 0){ i++; while(i < s.length() && s.charAt(i) == t.charAt(i)){ i++; } return i == s.length(); }else{ if(i == s.length()){ return true; }else{ while(i < s.length() && s.charAt(i) == t.charAt(i + 1)){ i++; } return i == s.length(); } } } }
又观察了一下,其实shift == 1 的情况下没有必要单独讨论 if(i == s.length()),可以整合到下面的判断中, 改进如下
public class Solution {
public boolean isOneEditDistance(String s, String t) {
// make s to be the shorter String
if(s.length() > t.length()){
return isOneEditDistance(t, s);
}
if(t.length() - s.length() > 1){
return false;
}
int shift = t.length() - s.length();
int i = 0;
while(i < s.length() && s.charAt(i) == t.charAt(i)){
i++;
}
if(shift == 0){
i++;
while(i < s.length() && s.charAt(i) == t.charAt(i)){
i++;
}
return i == s.length();
}else{
while(i < s.length() && s.charAt(i) == t.charAt(i + 1)){
i++;
}
return i == s.length();
}
}
}
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