闯关leetcode——66. Plus One

news2024/12/24 9:12:29

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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;
    }
};