[LeetCode] 66. Plus One

Posted 2021-01-03 arcsinw

Given a non-empty array of digits representing a non-negative integer, plus one to the integer.

The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.

You may assume the integer does not contain any leading zero, except the number 0 itself.

Example 1:

Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.

Example 2:

Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.

---

把一个vector存储的digits当成int来做加法，只+1

和[LeetCode] 2. Add Two Numbers有点像，然后我写出了一样（丑）的代码

从digits的末尾开始遍历，末尾+1，记下进位carry，如果没进位就直接返回digits，否则就继续看下一位，把进位传递下去

vector<int> plusOne(vector<int>& digits) {
    ++digits[digits.size()-1];
    int carry = digits[digits.size()-1] / 10;
    digits[digits.size()-1] %= 10;

    if (carry == 0)
    {
        return digits;
    }

    for(int i = digits.size()-2; i >= 0; i--)
    {
        digits[i] += carry;
        carry = digits[i] / 10;
        digits[i] %= 10;
    }

    if (carry != 0)
    {
        digits.insert(digits.begin(), 1);
    }

    return digits;
}

再看看LeetCode上其他代码，很简洁，但不是我现在能想出来的代码

由于是+1，digits[i]进位完后肯定是0，carry为1，所以循环里直接digits[i]++

vector<int> plusOne(vector<int> &digits) {
    for(int i = digits.size() - 1; i >= 0; i --){
        digits[i] ++;
        if(digits[i] < 10)
            break;
        digits[i] = 0;
    }
    if(digits[0] == 0){ // 99 + 1 = 100这种
        digits[0] = 1;
        digits.push_back(0);
    }
    return digits;
}