topic
leetcode66. Add one
Given a non-negative integer represented by a non-empty array of integers, add one to the number.
The highest digit is stored in the first position of the array, and each element in the array only stores a single number.
You can assume that in addition to the integer 0, this integer does not start with a zero.
Example 1:
Input: digits = [1,2,3]
Output: [1,2,4]
Explanation: The input array represents the number 123.
Example 2:
Input: digits = [4,3,2,1]
Output: [4,3,2,2]
Explanation: The input array represents the number 4321.
Example 3:
Input: digits = [0]
Output: [1]
提示：
1 <= digits.length <= 100
0 <= digits[i] <= 9
Code
- C code
- Ideas:
- The C code involves malloc, so you need to confirm the size of the requested resource first.
- Only when the numbers are all 9, returnSize = digitsSize + 1, otherwise returnSize = digitsSize.
- In the first case, the first digit of the result is 1, and the other digits are all 0, and the result can be returned directly.
- In the second case, just traverse the processing from the end to the front. The logic of addition: use modulo for standard position and divide for carry.
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
#include <malloc.h>
#include <string.h>
int* plusOne(int* digits, int digitsSize, int* returnSize){
*returnSize = digitsSize;
// 只有全9才会比digitsSize多一位
bool need_one_more = true;
for (int i = 0; i < digitsSize; ++i){
if (9 != digits[i]) {
need_one_more = false;
break;
}
}
if (need_one_more){
*returnSize = digitsSize + 1;
}
int* res = (int*)malloc(sizeof(int) * (*returnSize));
// 第1种情况
if (need_one_more){
res[0] = 1;
for (int i = 1; i < *returnSize; ++i){
res[i] = 0;
}
return res;
}
// 第2种情况
int temp = 1;
for (int i = digitsSize - 1; i >=0; --i){
res[i] = (digits[i] + temp) % 10;
temp = (digits[i] + temp) / 10;
}
return res;
}
- C++ code
- Does not involve resource application issues, just use pre-insertion in the vector.
#include <vector>
using namespace std;
class Solution {
public:
vector<int> plusOne(vector<int>& digits) {
int temp = 1;
vector<int> res;
for(auto it = digits.rbegin(); it != digits.rend(); ++it){
res.emplace(res.begin(), (*it + temp) % 10);
temp = (*it + temp) / 10;
}
// 注意最后可能的进位
if (temp){
res.emplace(res.begin(), temp);
}
return res;
}
};
test
#include <iostream>
template <typename C>
void print(C c, int size, char* s) {
cout << s << ": ";
for (int i = 0; i < size; ++i){
cout << c[i] << " ";
}
cout << endl;
}
int main() {
{
int digits[] = {
9};
int n = sizeof(digits) / sizeof(digits[0]);
print(digits, n, "digits");
int size;
int* output = plusOne(digits, n, &size);
print(output, size, "output");
}
{
vector<int> digits = {
9};
print(digits, digits.size(), "digits");
Solution s;
auto output = s.plusOne(digits);
print(output, output.size(), "output");
}
cin.get();
return 0;
}
- result
digits: 9
output: 1 0
digits: 9
output: 1 0