LeetCode 66 questions, plus one

Topic Overview

• Title: buckle force: 66 plus one
• Difficulty: Easy
• content:

  Given a non-negative integer nonempty array of integers represented, on the basis of the number plus one.

  Most significant digit stored in the first array, each element of the array stores only a single digit.

  You may assume that in addition to the integer 0, the integer does not begin with a zero.

  Example 1:

  Input: [1,2,3] Output: [2,4] Explanation: numeral 123 denotes the input array.

  Example 2:

  Input: [4,3,2,1] Output: [4,3,2,2] Explanation: input array represents a number 4321.

  Source: stay button (LeetCode) link: https: //leetcode-cn.com/problems/plus-one

The first idea

While loop becomes an integer array, and then calculating the num number of digits, then backwards cycle, integer array becomes.

Code

class Solution {
public:
    vector<int> plusOne(vector<int>& digits) {
       int num=0;
       int i=0;
       int a;
       for(int i(0);i<digits.size();i++){
           num=num*10+digits[i];
       } 
       num++;
       while(num/10!=0){
        i++;
       }
       do{
          a= num%10;
          digits[i]=a;
          i--;
          num=num/10;
       }while(i==0);    
       return digits;
    }
};

Test Submit

analysis

Running time is too long

Improve

Re-organize your thoughts:
total is divided into three cases:
. ① array 9 is not the last one, will not carry the plus one, such as: [1,2,3], plus a still three
. ② array final there are nine, but the first one is not 9, plus a still does not change the number of elements in the array, such as [1,9,9,9]
③. array book 9, add one would add an element, such as [9 , 9,9]
solution:
. ① long time to find reverse circulation 9 becomes 0, until the last one in a row of 9, plus one to the previous one element
② last case instead of the first. element to 0, the last one plus a 0;

Improved Code

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

Improved Submit

Harvest summary

push_back is a programming language function name inside. As the vector c ++ header file push_back which have this function, as added to the tail of vector data in a vector type in effect.