66. Plus One

置顶

Created At : 2021-04-07 23:58

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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.

将一个数字的每个位上的数字分别存到一个一维向量中，最高位在最开头，我们需要给这个数字加一，即在末尾数字加一，如果末尾数字是9，那么则会有进位问题，而如果前面位上的数字仍为9，则需要继续向前进位。具体算法如下：首先判断最后一位是否为9，若不是，直接加一返回，若是，则该位赋0，再继续查前一位，同样的方法，直到查完第一位。如果第一位原本为9，加一后会产生新的一位，那么最后要做的是，查运算完的第一位是否为0，如果是，则在最前头加一个1。代码如下：

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;
    }
}

我们也可以使用跟之前那道 Add Binary 类似的做法，将 carry 初始化为1，然后相当于 digits 加了一个0，处理方法跟之前那道题一样，参见代码如下：

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

类似题目：

Add Binary

Multiply Strings

Plus One Linked List

Add to Array-Form of Integer

Minimum Operations to Reduce an Integer to 0

参考资料：

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)