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66. Plus One

时间:2016-06-15 23:36:02      阅读:121      评论:0      收藏:0      [点我收藏+]

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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

66. Plus One

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

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