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 <= A.length <= 10000
0 <= A[i] <= 9
0 <= K <= 10000
- If
A.length > 1
, thenA[0] != 0
这题肯定有简单解法,写大数加法只是更熟练而已
class Solution { public: vector<int> addToArrayForm(vector<int>& A, int K) { int len = A.size(); int num=0; vector<int>Ve; int a[11000]={0},b[11000]={0},c[11000]={0}; for(int i=len-1;i>=0;i--){ a[num++]=A[i]; //cout<<A[i]<<endl; } int cnt=0; int i; while(K){ b[cnt++]=K%10; K/=10; } //cout<<"B"<<endl; int k=max(len,cnt); for(i=0;i<k;i++) { //cout<<a[i]<<" "<<b[i]<<endl; c[i]=a[i]+b[i]; } for(i=0; i<k; i++){ //cout<<"A1"<<endl; if(c[i]>=10){ c[i+1]+=c[i]/10; c[i]%=10; } //cout<<"A2"<<endl; } if(c[k]==0) k--; for(i=k;i>=0;i--){ Ve.push_back(c[i]); } return Ve; } };
原文地址:https://www.cnblogs.com/yinghualuowu/p/10360592.html
时间: 2024-10-21 09:36:01