• #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

     

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