You are given a 0-indexed integer array nums.
The concatenation of two numbers is the number formed by concatenating their numerals.
For example, the concatenation of 15, 49 is 1549.
The concatenation value of nums is initially equal to 0. Perform this operation until nums becomes empty:
If there exists more than one number in nums, pick the first element and last element in nums respectively and add the value of their concatenation to the concatenation value of nums, then delete the first and last element from nums.
If one element exists, add its value to the concatenation value of nums, then delete it.
Return the concatenation value of the nums.
Example 1:
Input: nums = [7,52,2,4]
Output: 596
Explanation: Before performing any operation, nums is [7,52,2,4] and concatenation value is 0.
Input: nums = [5,14,13,8,12]
Output: 673
Explanation: Before performing any operation, nums is [5,14,13,8,12] and concatenation value is 0.
Constraints:
1 <= nums.length <= 1000
1 <= nums[i] <= 104
我的垃圾解法
class Solution {
public:
int digits(int num){
int count = 0;
while(num){
num /= 10;
count++;
}
return count;
}
long long findTheArrayConcVal(vector<int>& nums) {
int n = nums.size();
if(n == 0) return 0;
else if(n == 1) return nums[0];
int i = 0,j = n - 1;
int tmp1 = nums[i], tmp2 = nums[j];
int tmp = tmp1 * pow(10,digits(tmp2)) + tmp2;
nums.erase(nums.begin() + j);
nums.erase(nums.begin() + i);
return tmp + findTheArrayConcVal(nums);
}
};
官方题解的解法
使用stoi函数和to_string函数求concatenation的值,更快捷
class Solution {
public:
long long findTheArrayConcVal(vector<int>& nums) {
long long ans = 0;
for (int i = 0, j = nums.size() - 1; i <= j; i++, j--) {
if (i != j) {
ans += stoi(to_string(nums[i]) + to_string(nums[j]));
} else {
ans += nums[i];
}
}
return ans;
}
};