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简单
数组
数学

English Version

题目描述

整数的 数组形式  num 是按照从左到右的顺序表示其数字的数组。

  • 例如,对于 num = 1321 ,数组形式是 [1,3,2,1]

给定 num ,整数的 数组形式 ,和整数 k ,返回 整数 num + k数组形式

 

示例 1:

输入:num = [1,2,0,0], k = 34
输出:[1,2,3,4]
解释:1200 + 34 = 1234

示例 2:

输入:num = [2,7,4], k = 181
输出:[4,5,5]
解释:274 + 181 = 455

示例 3:

输入:num = [2,1,5], k = 806
输出:[1,0,2,1]
解释:215 + 806 = 1021

 

提示:

  • 1 <= num.length <= 104
  • 0 <= num[i] <= 9
  • num 不包含任何前导零,除了零本身
  • 1 <= k <= 104

解法

方法一

Python3

class Solution:
    def addToArrayForm(self, num: List[int], k: int) -> List[int]:
        i, carry = len(num) - 1, 0
        ans = []
        while i >= 0 or k or carry:
            carry += (0 if i < 0 else num[i]) + (k % 10)
            carry, v = divmod(carry, 10)
            ans.append(v)
            k //= 10
            i -= 1
        return ans[::-1]

Java

class Solution {
    public List<Integer> addToArrayForm(int[] num, int k) {
        int i = num.length - 1, carry = 0;
        LinkedList<Integer> ans = new LinkedList<>();
        while (i >= 0 || k > 0 || carry > 0) {
            carry += (i < 0 ? 0 : num[i--]) + k % 10;
            ans.addFirst(carry % 10);
            carry /= 10;
            k /= 10;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> addToArrayForm(vector<int>& num, int k) {
        int i = num.size() - 1, carry = 0;
        vector<int> ans;
        for (; i >= 0 || k || carry; --i) {
            carry += (i < 0 ? 0 : num[i]) + k % 10;
            ans.push_back(carry % 10);
            carry /= 10;
            k /= 10;
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

Go

func addToArrayForm(num []int, k int) []int {
	i, carry := len(num)-1, 0
	ans := []int{}
	for ; i >= 0 || k > 0 || carry > 0; i-- {
		if i >= 0 {
			carry += num[i]
		}
		carry += k % 10
		ans = append(ans, carry%10)
		carry /= 10
		k /= 10
	}
	for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
		ans[i], ans[j] = ans[j], ans[i]
	}
	return ans
}

TypeScript

function addToArrayForm(num: number[], k: number): number[] {
    let arr2 = [...String(k)].map(Number);
    let ans = [];
    let sum = 0;
    while (num.length || arr2.length || sum) {
        let a = num.pop() || 0,
            b = arr2.pop() || 0;
        sum += a + b;
        ans.unshift(sum % 10);
        sum = Math.floor(sum / 10);
    }
    return ans;
}

Rust

impl Solution {
    pub fn add_to_array_form(num: Vec<i32>, mut k: i32) -> Vec<i32> {
        let n = num.len();
        let mut res = vec![];
        let mut i = 0;
        let mut sum = 0;
        while i < n || sum != 0 || k != 0 {
            sum += num.get(n - i - 1).unwrap_or(&0);
            sum += k % 10;
            res.push(sum % 10);

            i += 1;
            k /= 10;
            sum /= 10;
        }
        res.reverse();
        res
    }
}

方法二

TypeScript

function addToArrayForm(num: number[], k: number): number[] {
    const n = num.length;
    const res = [];
    let sum = 0;
    for (let i = 0; i < n || sum !== 0 || k !== 0; i++) {
        sum += num[n - i - 1] ?? 0;
        sum += k % 10;
        res.push(sum % 10);
        k = Math.floor(k / 10);
        sum = Math.floor(sum / 10);
    }
    return res.reverse();
}