Leetcode:one Edit Distance

if They is both one edit distance apart.

With I, j two pointers, sweep from left to right, two go together.

S.charat (i)! = T.charat (j)

distance++

Either I jump one cell, J jumps a grid, or i,j jumps together. Corresponding to: INSERT, delete, replace.

Then if distance >1, return false;

The base case has three:

1. I, J all go to their respective string.length (), because as long as the dist>1 will return false, so there is certainly no dist>1, so the direct return to True

2. I go to the far right of S, J has not, so the current dist to add t.length ()-j this paragraph.

3. J go to the far right of T, I have not, so the current dist to add s.length ()-I this paragraph.

Did not do OJ do not know whether will be over, O (N) scan once, not calculate recursion stack space complexity O (1)

1  Public classSolution {2     Public Booleanisoneeditdistance (string s, String t) {3        if(s==NULL|| t==NULL)return false;4        returnCheck (s, t, 0, 0, 0);5    }6    7     Public BooleanCheck (string s, String T,intIintJintDist) {8        if(i = = S.length () && j = t.length ())return true;9        Else if(i = =s.length ()) {TenDist + = T.length ()-J; One            returnDist>1?false:true; A        } -        Else if(J = =t.length ()) { -Dist + = s.length ()-i; the            returnDist>1?false:true; -        } -        if(S.charat (i)! =T.charat (j)) { -dist++; +            if(Dist > 1)return false; -            returnCheck (s, T, I+1, J, dist) | | Check (s, T, I, j+1, dist) | | Check (s, T, I+1, j+1, Dist); +        } A        Else returnCheck (s, T, I+1, j+1, Dist); at    } -}

Other people's approach:

1  Public classSolution {2     Public Booleanisoneeditdistance (string s, String t) {3        returnCheck (s,t,0,0,0);4   }5   6    Public BooleanCheck (string s, String T,intIintJintdistance) {7        while(0<=i && i<=s.length ()-1 && 0<=j && j<=t.length ()-1){8           if(S.charat (i)! =T.charat (j)) {9distance++;Ten               if(distance>1)return false; One               returnCheck (s,t,i+1,j,distance) | | Check (s,t,i,j+1,distance) | | Check (s,t,i+1,j+1, distance); A}Else{ -               returnCheck (s,t,i+1,j+1, distance); -           } the       } -        -       if(Distance ==1){ -           returnI==s.length () && j==t.length (); +}Else{//(distance ==0) -           returnMath.Abs (S.length ()-t.length ()) ==1; +       } A   } at}

In addition, this problem is similar to edit distance, can be done with that method, but will tle

1  Public classSolution {2     Public Booleanisoneeditdistance (string s, String t) {3        if(s==NULL|| t==NULL)return false;4        int[] res =New int[S.length () +1] [T.length () +1];5Res[0][0] = 0;6         for(intI=1; I<=s.length (); i++) {7Res[i][0] =i;8        }9         for(intJ=1; J<=t.length (); J + +) {TenRES[0][J] =J; One        } A         for(intI=1; I<=s.length (); i++) { -             for(intJ=1; J<=t.length (); J + +) { -                intMinoftwo = Math.min (Res[i-1][j], res[i][j-1]) + 1; the                intReplace = S.charat (i-1) = = T.charat (j-1)? RES[I-1][J-1]: res[i-1][j-1]+1; -RES[I][J] =math.min (minoftwo, replace); -                if(Res[i][j] >= 2)return false; -            } +        } -        return true; +    } A}

Leetcode:one Edit Distance