Problem:
Given two ints, each in the range 10..99, return true if there is a digit that appears in both numbers, such as the 2 in 12 and 23. (Note: division, e.g. n/10, gives the left digit while the % "mod" n%10 gives the right digit.)
shareDigit(12, 23) → true
shareDigit(12, 43) → false
shareDigit(12, 44) → false
Solution:
public boolean shareDigit(int a, int b) {
int aL = a / 10;
int aR = a % 10;
int bL = b / 10;
int bR = b % 10;
if (aL == bL || aL == bR || aR == bL || aR == bR)
return true;
else
return false;
}
int p,q,r=b;
ReplyDeletewhile(a!=0)
{
p=a%10;
while(b!=0)
{
q=b%10;
if(p==q)
return true;
b=b/10;
}
a=a/10;
b=r;
}
return false;
public boolean shareDigit(int a, int b) {
ReplyDeletereturn a/10==b/10||
a/10==b%10||
a%10==b%10||
a%10==b/10;
}
public boolean shareDigit(int a, int b) {
ReplyDeletereturn (a /10 == b /10 || a / 10 == b % 10) || (a %10 == b /10 || a % 10 == b % 10);
}
public boolean shareDigit(int a, int b) {
ReplyDeleteif(a/10==b/10||a/10==b%10)
{
return true;
}
else if((a%10==b%10||a%10==b/10))
{
return true;
}
return false;
}