Find no with even no of digits

.java

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+// Given an array nums of integers, return how many of them contain an even number of digits.
+
+// Example 1:
+// Input: nums = [12,345,2,6,7896]
+// Output: 2
+// Explanation: 
+// 12 contains 2 digits (even number of digits). 
+// 345 contains 3 digits (odd number of digits). 
+// 2 contains 1 digit (odd number of digits). 
+// 6 contains 1 digit (odd number of digits). 
+// 7896 contains 4 digits (even number of digits). 
+// Therefore only 12 and 7896 contain an even number of digits.
+// Example 2:
+// Input: nums = [555,901,482,1771]
+// Output: 1 
+// Explanation: 
+// Only 1771 contains an even number of digits.
+
+// Constraints:
+// 1 <= nums.length <= 500
+// 1 <= nums[i] <= 105
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+
+// Approach 1
+public int findNumbers(int[] nums) {
+        int count = 0, evenNo = 0;
+        for(int i = 0; i < nums.length; i++){
+           int j = nums[i];
+            while(j > 0){
+                count++;
+                j/= 10;
+            }
+          if(count %2 == 0)
+              evenNo++;
+            count = 0;
+        }
+      return evenNo;
+    }
+
+// Approach 2
+    public int findNumbers(int[] nums){
+        int even = 0;
+       for(int i = 0; i < nums.length; i++){
+           String a = String.valueOf(nums[i]);
+           if(a.length()%2 == 0)
+               even++;
+       }
+        return even;
+	}
+
+
+// Approach 3
+
+public int findNumbers(int[] nums){
+        int even = 0;
+        for(int i = 0; i < nums.length; i++){
+            if(nums[i]>9 && nums[i]<100 || nums[i]>999 && nums[i]<10000 || nums[i] == 100000)
+                even++;
+        } 
+        return even;
+    }
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