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66. Plus One

Easy
Description

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

解題思路

討論區有個好心人留了這段

Guys!! Try solving the problem using 3 cases.
a) Considering all the values are 9.
b) Considering the last digit is not 9.
c) Considering the last digit is 9.
This will make the coding a bit easy for you

所以這題可以簡化為以下三種狀況

  1. 最後一個數字是 9
  2. 最後一個數字不是 9
  3. 所有數字都是 9

如果最後一個數字是 9 的話,就要把這個數字設為 0,並且把下一個數字加一。
如果最後一個數字不是 9 的話,那就直接把最後一個數字加一就好。
而有可能會有連續都是 9 的情況,所以要用 for 迴圈來處理。

轉換為程式碼就會如下所示

var plusOne = function (digits) {
for (let i = digits.length - 1; i >= 0; i--) {
// 從最尾端開始
if (digits[i] === 9) {
// 如果是 9 的話
digits[i] = 0; // 就把這個數字設為 0,接著進到下一個 for 迴圈的下個階段
} else {
digits[i] += 1; // 如果不是 9 的話,就把這個數字加一
return digits; // 然後直接跳出迴圈回傳結果
}
}

digits.unshift(1); // 運行到這裡就表示 for 迴圈跑完,代表所有數字都是 9 ,要在最前面補上 1
return digits;
};

心得

2026 年的第一題 LeetCode 每日挑戰,新年新哩扣!