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  • 66. Plus One

    1. 问题描述

    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.

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    Tags: Array Math
    Similar Problems: (M) Multiply Strings (E) Add Binary

    2. 解答思路

    3. 代码

     1 #include <vector>
     2 using namespace std;
     3 class Solution {
     4 public:
     5     vector<int> plusOne_method_1(vector<int>& digits) {
     6         if (digits.empty())
     7         {
     8             digits.push_back(1);
     9             return digits;
    10         }
    11         return puls(digits, digits.size()-1, 1);
    12     }
    13     vector<int> plusOne_method_2(vector<int>& digits) {
    14         if (digits.empty())
    15         {
    16             digits.push_back(1);
    17             return digits;
    18         }
    19         int nTakeOver = 1;
    20         int i;
    21         for (i=digits.size()-1; i>=0; i--)
    22         {
    23             int sum = digits[i] + nTakeOver;
    24             digits[i] = sum%10;
    25             nTakeOver = sum/10;
    26             if (0 == nTakeOver)
    27             {
    28                 return digits;
    29             }
    30         }
    31         if (1 == nTakeOver)
    32         {
    33             digits.insert(digits.begin(), 1);
    34         }
    35         return digits;
    36     }
    37 private:
    38     vector<int> puls(vector<int>& digits, int idx, int nTakeOver = 0)
    39     {
    40         int sum = digits[idx] + nTakeOver;
    41         digits[idx] = sum%10;
    42         nTakeOver = sum/10;
    43         if (0 == nTakeOver)
    44         {
    45             return digits;
    46         }
    47         if (0 == idx)
    48         {
    49             digits.insert(digits.begin(), 1);
    50             return digits;
    51         }
    52         return puls(digits, idx-1, 1);
    53     }
    54 };

    4. 反思

    方法1. 递归实现

    方法2. 循环实现

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  • 原文地址:https://www.cnblogs.com/whl2012/p/5589259.html
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