算法:加一66. Plus One
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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 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].
Constraints:
- 1 <= digits.length <= 100
- 0 <= digits[i] <= 9
- digits does not contain any leading 0’s.
O(n) 遍历法
class Solution {
public int[] plusOne(int[] digits) {
int len = digits.length;
for (int i = len - 1; i >= 0; i--) {
if (digits[i] < 9) {
digits[i]++;
return digits;
}
digits[i] = 0;
}
int[] newDigits = new int[digits.length + 1];
newDigits[0] = 1;
return newDigits;
}
}
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