大纲
- 题目
- 地址
- 内容
- 解题
- 代码地址
题目
地址
https://leetcode.com/problems/plus-one/description/
内容
You are given a large integer represented as an integer array digits, where each digits[i] is the ith digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading 0’s.
Increment the large integer by one and return the resulting array of digits.
Example 1:
Input: digits = [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Incrementing by one gives 123 + 1 = 124.
Thus, the result should be [1,2,4].
Example 2:
Input: digits = [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
Incrementing by one gives 4321 + 1 = 4322.
Thus, the result should be [4,3,2,2].
Example 3:
Input: digits = [9]
Output: [1,0]
Explanation: The array represents the integer 9.
Incrementing by one gives 9 + 1 = 10.
Thus, the result should be [1,0].
Constraints:
- 1 <= digits.length <= 100
- 0 <= digits[i] <= 9
- digits does not contain any leading 0’s.
解题
这题要求给大数自增1。在64位系统中,一个整数的上限是264。在这个范围内([0,264))的整数都可以通过汇编指令直接自增。但是超过这个范围的数字,就需要借助其他方法了。本题将数字放在一个数组中,然后我们像做竖式计算一样,从后向前计算出结果。唯一需要额外考虑的是,如果最前面一位有进位,则需要在vector头部插入1这个新元素。
#include <vector>
using namespace std;
class Solution {
public:
vector<int> plusOne(vector<int>& digits) {
int carry = 1;
for (int i = digits.size() - 1; i >= 0; i--) {
int sum = digits[i] + carry;
digits[i] = sum % 10;
carry = sum / 10;
}
if (carry > 0) {
digits.insert(digits.begin(), carry);
}
return digits;
}
};
代码地址
https://github.com/f304646673/leetcode/tree/main/66-Plus-One