66. Plus One

1. Question

You are given a large integer represented as an integer array digits, where each digits[i] is the i^th 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.

2. Examples

Example 1:

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].

Example 2:

Input: digits = [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
Incrementing by one gives 4321 + 1 = 4322.
Thus, the result should be [4,3,2,2].

Example 3:

Input: digits = [0]
Output: [1]
Explanation: The array represents the integer 0.
Incrementing by one gives 0 + 1 = 1.
Thus, the result should be [1].

Example 4:

Input: digits = [9]
Output: [1,0]
Explanation: The array represents the integer 9.
Incrementing by one gives 9 + 1 = 10.
Thus, the result should be [1,0].

3. Constraints

• 1 <= digits.length <= 100
• 0 <= digits[i] <= 9
• digits does not contain any leading 0's.

4. References

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/plus-one 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

5. Solutions

主要考虑进位问题。需要双重遍历，第一次逆序找出总共需要进位的位数，第二次修改最高位自增并置后面的数为0.如果都是9，就新增一个空间。

class Solution {
  public int[] plusOne(int[] digits) {
    for (int i = digits.length - 1; i >= 0; i--) {
      if (digits[i] != 9) {
        digits[i]++;
        for (int j = i + 1; j < digits.length; j++) {
          digits[j] = 0;
        }
        return digits;
      }
    }
    int[] arr = new int[digits.length + 1];
    arr[0] = 1;
    return arr;
  }
}