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[LintCode] Plus One 加一运算

更新时间:2022-09-15 14:56:32

Given a non-negative number represented as an array of digits, plus one to the number.

The digits are stored such that the most significant digit is at the head of the list.

Example

Given [1,2,3] which represents 123, return[1,2,4].

Given [9,9,9] which represents 999, return[1,0,0,0].

LeetCode上的原题,请参见我之前的博客Plus One

解法一:

class Solution {
public:
    /**
     * @param digits a number represented as an array of digits
     * @return the result
     */
    vector<int> plusOne(vector<int>& digits) {
        vector<int> res;
        int carry = 1, n = digits.size();
        for (int i = n - 1; i >= 0; --i) {
            int sum = digits[i] + carry;
            res.insert(res.begin(), sum % 10);
            carry = sum / 10;
        }
        if (carry == 1) res.insert(res.begin(), 1);
        return res;
    }
};

解法二:

class Solution {
public:
    /**
     * @param digits a number represented as an array of digits
     * @return the result
     */
    vector<int> plusOne(vector<int>& digits) {
        for (int i = digits.size() - 1; i >= 0; --i) {
            if (digits[i] < 9) {
                digits[i] += 1;
                return digits;
            }
            digits[i] = 0;
        }
        digits.insert(digits.begin(), 1);
        return digits;
    }
};

本文转自博客园Grandyang的博客,原文链接:加一运算[LintCode] Plus One ,如需转载请自行联系原博主。