Given a non-empty array of digits representing a non-negative integer, plus one to the integer.

The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.

You may assume the integer does not contain any leading zero, except the number 0 itself.

Example 1:

Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Example 2:

Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/plus-one
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学了一个新语法，在vector前面加东西

digits.insert(digits.begin(),1);

class Solution {
public:
    vector<int> plusOne(vector<int>& digits) {
        if(digits.size()<1){
            digits.push_back(1);
            return digits;
        }digits[digits.size()-1]+=1;
        for(int i=digits.size()-1;i>0;i--){
            if(digits[i]>9){
                digits[i-1]++;
                digits[i]-=10;
            }else{
                break;
            }
        }if(digits[0]>9){
            digits[0]-=10;
            digits.insert(digits.begin(),1);
                         }
        return digits;
        
    }
};

最快的那个人这么想的~~~~

因为如果要进位，一定是  99 999 9999 这种，所以就全变成了0，直接第一个变成1，尾部加0效果是一样的~~~厉害了

class Solution {
public:
    vector<int> plusOne(vector<int>& digits) {
        for (int i = (int)digits.size() - 1; i >= 0; i--) {
            if (digits[i] == 9) {
                digits[i] = 0;
            }
            else {
                digits[i]++;
                break;
            }
        }
        if (digits[0] == 0) {
            digits.push_back(0);
            digits[0] = 1;
        }
        return digits;
    }
};