Plus One
LeetCode.66.Plus One You are given a large integer represented as an integer array digits, where each digits[i] is the ith digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading 0’s.
Increment the large integer by one and return the resulting array of digits.
Example:
Input: digits = [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Incrementing by one gives 123 + 1 = 124.
Thus, the result should be [1,2,4].
First Solution
public static int[] intArray(int[] digits) {
int leng = digits.length;
StringBuffer stringBuffer = new StringBuffer();
for (int i = 0; i < leng; i++) {
stringBuffer.append(digits[i]);
}
Integer num = Integer.parseInt(stringBuffer.toString()) + 1;
String str = num.toString();
int[] intArray = new int[str.length()];
for (int i = 0; i < str.length(); i++) {
Character ch = str.charAt(i);
intArray[i] = Integer.parseInt(ch.toString());
}
return intArray;
}
When digits=[9,8,7,6,5,4,3,2,1,0], it will appear NumberFormatException
Other Solution
public int[] plusOne(int[] digits) {
for (int i = digits.length - 1; i >= 0; i--) {
if (digits[i] != 9) {
digits[i]++;
return digits;
}
digits[i] = 0;
}
int[] temp = new int[digits.length + 1];
temp[0] = 1;
return temp;
}