Some ideas About invisible bugs
Leetcode 532 as example:
Following is the right solution:

class Solution {
    public int findPairs(int[] nums, int k) {
        int res = 0;
        Arrays.sort(nums);
        int i = 0;
        int j = 0;
        while (i < nums.length && j < nums.length) {
            j = Math.max(j, i + 1); //each time, j will relocate to max(j, i+1)
            while (j < nums.length && nums[j] - nums[i] < k) { //j will move to the place where nums[j]-nums[i]>=k
                j++;
            }
            if (j < nums.length && nums[j] - nums[i] == k) {
                res++;
            }
            while (i + 1 < nums.length && nums[i] == nums[i + 1]) { //i will move to the position where last duplicate exists
                i++;
            }
            i++;
        }
        return res;
    }
}

Following is my answer, which only differs with the correct solution in inner while loops(mu solution is while-while-if and the correct solution is while0if-while)
But my answer can’t pass all test cases, beside that, everything is exactly the same.

class Solution {
    public int findPairs(int[] nums, int k) {
        if (nums == null || nums.length < 2) {
            return 0;
        }
        
        Arrays.sort(nums);
        int i = 0;
        int j = 0;
        int count = 0;
        while (i < nums.length && j < nums.length) {
            while (i < nums.length - 1 && nums[i] == nums[i+1]) {
                i++;
            } //i will be moved to the index of last duplicate
            j = Math.max(j, i + 1);
            while (j < nums.length && nums[j] - nums[i] < k) { //we can use binary seach to find such target in that part sort list
                j++;
            } 
            if (j < nums.length && nums[j] - nums[i] == k) {
                count++;
            }//else nothing
            i++;
        }
        return count;
    }
}

and we can’t use .clear() method, if we want to reuse a list. like code below:

// Java code to illustrate clear() method 
import java.io.*; 
import java.util.*; 

public class ListDemo { 
	public static void main(String[] args) 
	{ 

		// create an empty list with an initial capacity 
		List<Integer> list = new ArrayList<Integer>(5); 

		// use add() method to initially 
		// add elements in the list 
		list.add(10); 
		list.add(20); 
		list.add(30); 
        List<List<Integer>> test = new ArrayList<>();
      test.add(list);

		// clear the list 
		list.clear(); 
      list.add(1);
        List<List<Integer>> test1 = new ArrayList<>();
      test1.add(list);

		// prints all the elements available in list 
		System.out.println(test);
      System.out.println(test1);
	} 
} 
//and the test and tes1 list are both print out [1]