# 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 = [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.

C++ 解法一：

``````class Solution {
public:
vector<int> plusOne(vector<int>& digits) {
int n = digits.size();
for (int i = n - 1; i >= 0; --i) {
if (digits[i] == 9) digits[i] = 0;
else {
digits[i] += 1;
return digits;
}
}
if (digits.front() == 0) digits.insert(digits.begin(), 1);
return digits;
}
};
``````

Java 解法一：

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

C++ 解法二 ：

``````class Solution {
public:
vector<int> plusOne(vector<int>& digits) {
int carry = 1, n = digits.size();
for (int i = n - 1; i >= 0; --i) {
if (carry == 0) return digits;
int sum = digits[i] + carry;
digits[i] = sum % 10;
carry = sum / 10;
}
if (carry == 1) digits.insert(digits.begin(), 1);
return digits;
}
};
``````

Java 解法二 ：

``````public class Solution {
public int[] plusOne(int[] digits) {
int carry = 1, n = digits.length;
for (int i = digits.length - 1; i >= 0; --i) {
if (carry == 0) return digits;
int sum = digits[i] + carry;
digits[i] = sum % 10;
carry = sum / 10;
}
int[] res = new int[n + 1];
res[0] = 1;
return carry == 0 ? digits : res;
}
}
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/66

Multiply Strings

Minimum Operations to Reduce an Integer to 0

https://leetcode.com/problems/plus-one/

https://leetcode.com/problems/plus-one/discuss/24082/my-simple-java-solution

https://leetcode.com/problems/plus-one/discuss/24084/Is-it-a-simple-code(C%2B%2B)

LeetCode All in One 题目讲解汇总(持续更新中…)

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