#Leetcode# 989. Add to Array-Form of Integer

https://leetcode.com/problems/add-to-array-form-of-integer/

For a non-negative integer X, the array-form of X is an array of its digits in left to right order.  For example, if X = 1231, then the array form is [1,2,3,1].

Given the array-form A of a non-negative integer X, return the array-form of the integer X+K.

Example 1:

Input: A = [1,2,0,0], K = 34
Output: [1,2,3,4]
Explanation: 1200 + 34 = 1234

Example 2:

Input: A = [2,7,4], K = 181
Output: [4,5,5]
Explanation: 274 + 181 = 455

Example 3:

Input: A = [2,1,5], K = 806
Output: [1,0,2,1]
Explanation: 215 + 806 = 1021

Example 4:

Input: A = [9,9,9,9,9,9,9,9,9,9], K = 1
Output: [1,0,0,0,0,0,0,0,0,0,0]
Explanation: 9999999999 + 1 = 10000000000

Note:

  1. 1 <= A.length <= 10000
  2. 0 <= A[i] <= 9
  3. 0 <= K <= 10000
  4. If A.length > 1, then A[0] != 0

代码:

class Solution {
public:
    vector<int> addToArrayForm(vector<int>& A, int K) {
        int n = A.size();
        vector<int> ans;
        string words1 = toString(K), words2 = "";
        for(int i = 0; i < n; i ++)
            words2 += ('0' + A[i]);
        string sum = addStrings(words1, words2);
        for(int i = 0; i < sum.length(); i ++)
            ans.push_back(sum[i] - '0');
        return ans;
    }
    string toString(int x) {
        string s = "";
        while(x) {
            s += (x % 10) + '0';
            x /= 10;
        }
        for(int i = 0; i < s.length() / 2; i ++)
            swap(s[i], s[s.length() - i - 1]);
        return s;
    }
    string addStrings(string num1, string num2) {
        string c = "";
        int len1 = num1.length();
        int len2 = num2.length();
        int len = max(len1, len2);
        for(int i = len1; i < len; i ++)
            num1 = "0" + num1;
        for(int i = len2; i < len; i ++)
            num2 = "0" + num2;
        int ok = 0;
        for(int i = len - 1; i >= 0; i --) {
            char temp = num1[i] + num2[i] - '0' + ok;
            if(temp > '9') {
                ok = 1;
                temp -= 10;
            }
            else ok = 0;
            c = temp + c;
        }
        if(ok) c = "1" + c;
        return c;
    }
};

  数组模拟大数加法

 

原文地址:https://www.cnblogs.com/zlrrrr/p/10691966.html