1295. Statistical even number of digits to digital
Subject description:
To give you an array of integers nums, you return to where the median of an even number of numbers.
Example 1:
Input: nums = [12,345,2,6,7896]
Output: 2
Explanation:
12 is 2 digits (number of bits is an even number)
345 3 digits (digits is odd)
2 is a 1-bit digital (odd digits )
6 is an odd number of digits)
7896 4 digits (number of bits is an even number)
so that only the 7896 is 12 bits and the even-numbered digital
example 2:
Input: nums = [555,901,482,1771]
Output: 1
Explanation:
Only median number 1771 is an even number.
prompt:
1 <= nums.length <= 500
1 <= nums[i] <= 10^5
Problem solution 1:
Title numbers are given size range 1 to 100,000, where the number of bits is an even number are:
* When one hundred thousand is not 0, a count is incremented; (100000)
· When ten thousand and one hundred thousand to 0, one thousand is not 0 when the count plus one; (001000)
when the current face are all 0 · ten bits are not 0, a count is incremented. (000010)
public class Solution2 {
public static int findNumbers(int[] nums) {
//统计位数为偶数的数字个数
int count = 0;
//该循环用于判断数字位数是否为偶数
for(int i = 0;i<nums.length;i++){
int n = nums[i];
int shiwan = n/100000;
int wan = n/10000;
int qian = n/1000;
int bai = n/100;
int shi = n/10;
if(shiwan!=0){
count++;
}else if(wan!=0){
}else if(qian!=0){
count++;
}else if(bai!=0){
}else if(shi!=0){
count++;
}
}
//返回个数
return count;
}
Submit Result 1:
Problem solution 2:
Converts the number into a string type, length is determined so that the code is simpler
class Solution {
public int findNumbers(int[] nums) {
//统计位数为偶数的数字个数
int count = 0;
for (int num : nums)
//判断转换成的字符串长度是否为偶数，是计数就加1
count += String.valueOf(num).length() % 2 == 0 ? 1 : 0;
return count;
}
}
2 to submit the results:
