LeetCode - Plus One

Plus One

2013.12.22 03:37

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

 

Solution:

　　This problem seems quite easy, right? See if you can AC with one shot.

　　Here're some good test cases for you:

　　　　0 + 1 = 1

　　　　1 + 1 = 2

　　　　34 + 1 = 35

　　　　99 + 1 = 100

　　Note that we write the number "100" in the order '1','0','0', it's stored as ['1', '0', '0']. Don't make it 001.

　　Time compelxity is O(n), where n is the length of the string.Space compelxity is O(1).

Accepted code:

 1 // 3WA, 1AC

 2 class Solution {

 3 public:

 4     vector<int> plusOne(vector<int> &digits) {

 5         // IMPORTANT: Please reset any member data you declared, as

 6         // the same Solution instance will be reused for each test case.

 7         

 8         int i, len;

 9         len = digits.size();

10         

11         if(len <= 0){

12             return digits;

13         }

14         

15         // 1WA here, add 1 at the wrong position, you FOOL!!!

16         ++digits[len - 1];

17         // 1WA here, reversed order..

18         for(i = len - 1; i > 0; --i){

19             digits[i - 1] += digits[i] / 10;

20             digits[i] %= 10;

21         }

22         // 1WA here, carry propagation here neglected..

23         if(digits[0] >= 10){

24             int tmp = digits[0] / 10;

25             digits[0] %= 10;

26             digits.push_back(0);

27             for(i = len; i > 0; --i){

28                 digits[i] = digits[i - 1];

29             }

30             digits[0] = tmp;

31         }

32         

33         return digits;

34     }

35 };