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法律学按照给定的约束添加给定数组中的元素?
这里解决一个问题。我们将两个数组的元素相加并存储在另一个数组中。不过，我们要遵循一些约束条件。这些约束条件如下所示 −
- 两个数组的第 0 个索引开始进行相加
- 如果和大于一位数，则将和拆分，并将每位数字放置在相应的位置
- 较大输入数组的剩余数字将存储在输出数组中
让我们看看算法以了解思路。
算法
addArrayConstraints(arr1, arr2)
Begin define empty vector out i := 0 while i is less than both arr1.length and arr2.length, do add := arr1[i] + arr2[i] if add is single digit number, then insert add into out else split the digits and push each digit into out end if done while arr1 is not exhausted, do insert each element directly into out if they are single digit, otherwise split and insert done while arr2 is not exhausted, do insert each element directly into out if they are single digit, otherwise split and insert done End
示例
#include<iostream>
#include<vector>
using namespace std;
void splitDigit(int num, vector<int> &out) { //split the digits and store into vector to add into array
vector<int> arr;
while (num) {
arr.insert(arr.begin(), num%10);
num = num/10;
}
out.insert(out.end(), arr.begin(), arr.end());
}
void addArrayConstraints(int arr1[], int arr2[], int m, int n) {
vector<int> out;
int i = 0; //point current index of arr1 and arr2
while (i < m && i <n) {
int add = arr1[i] + arr2[i];
if (add < 10) //if it is single digit, put the sum, otherwise split them
out.push_back(add);
else
splitDigit(add, out);
i++;
}
}
while (i < m) //if arr1 has more elements
splitDigit(arr1[i++], out);
while (i < n) //if arr2 has more elements
splitDigit(arr2[i++], out);
for (int i = 0; i< out.size(); i++)
cout << out[i] << " ";
}
main() {
int arr1[] = {9323, 8, 6, 55, 25, 6};
int arr2[] = {38, 11, 4, 7, 8, 7, 6, 99};
int n1 = sizeof(arr1) / sizeof(arr1[0]);
int n2 = sizeof(arr2) / sizeof(arr2[0]);
addArrayConstraints(arr1, arr2, n1, n2);
}输出
9 3 6 1 1 9 1 0 6 2 3 3 1 3 6 9 9
更新于: 2020 年 7 月 2 日
99 次浏览
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