Given a non-negative number represented as an array of digits,
add 1 to the number ( increment the number represented by the digits ).
The digits are stored such that the most significant digit is at the head of the list.
Example 1:
If the vector has [1, 2, 3]
The returned vector should be [1, 2, 4]
Reverse the digits of the number to make your life easier. Maintain a carry which is initialized to 1 ( Denoting that we need to add one to the number ). Run a loop on the digit array ( which is now reversed ), and add carry to the current digit. If the digit D exceeds 10 on doing so, store the last digit in current position ( D % 10 ), and make carry = rest of the digits ( D / 10 ). Continue the process till you have covered all the digits. Now if at the end, carry = 0, we are done, and we can return the array. If not, we need to add one more digit, with carry in it.
class Solution:
def plusOne(self,A):
val = 1;
for i in range(len(A),0,-1):
val = val + A[i-1]
borrow = int(val/10)
if borrow == 0:
A[i-1] = val
break;
else:
A[i-1] = val%10
val = borrow
A = [borrow] + A
while A[0] == 0:
del A[0]
return Aclass solution
{
public:
vector plusOne(vector& arr)
{
int n = arr.size();
arr[n-1] += 1;
int hand = arr[n-1]/10;
arr[n-1] = arr[n-1] % 10;
for (int i=n-2; i>=0; i--)
{
if (hand == 1)
{
arr[i] += 1;
hand = arr[i]/10;
arr[i] = arr[i] % 10;
}
}
if (hand == 1)
arr.insert(arr.begin(), 1);
return arr;
}
};