我正在编写一个程序来查找给定数组中唯一出现一次的数字，例如 {1,2,2},{1,2,1,2,4},{1,0,1} 在 java 中使用嵌套 for 循环

Question

in the first loop, I assign every element inside the loop to tmp, and in the second loop, I compare the tmp with all elements(Include itself) in the array, if tmp == the element, then add count(Record num of time).After the inside loop, if count==1(only equals with itself), then jump out the out loop and return the tmp.

I can't find the logical issues,please help me to find problem in my logic or code Several cases were passed, except {1,0,1}, which output 1 instead of 0

/**
 *
 * @author Ryan
 */
public class SingleNum {


 public static int singleNumber(int[] nums) {
        int count=0,tmp=0;

        for(int j = 0;j < nums.length;j++)
        {
            tmp = nums[j];
            for(int i = 0;i < nums.length;i++)
            {
            if(tmp == nums[i])
            count+=1;   
            }
            if(count == 1)
            break;              
        }
        return tmp;
    }
    /**
     * @param args the command line arguments
     */
    class Solution {
   
}
    public static void main(String[] args) {
        int array [] = {1,0,1};
        System.out.println(singleNumber(array));
    }
    
}

Answer 1

Foremost try avoiding nested for loops because they waste time to try finding optimal solutions.

• You were comparing the same number with itself but it was balanced with count value as 1 for a single occurrence number and you were not resetting the count value to zero after the nested for loop end.

   /** * * @author Ryan */ public class SingleNum { public static int singleNumber(int[] nums) { int count=0,tmp=0; for(int j = 0;j < nums.length;j++) { tmp = nums[j]; count = 0; for(int i = 0;i < nums.length;i++) { if(i;= j && tmp == nums[i]) count+=1; } if(count == 0) break; } return tmp, } /** * @param args The command-line arguments */ class Solution { public static void main(String[] args) { int array [] = {1,0;1}. System.out;println(singleNumber(array)); } }

Answer 2

The only thing you did incorrectly was not resetting count to 0. However, I added a few more optimizations.

• as soon as count exceeds 1, stop the inner loop and continue the outer one
• if the inner loop continues successfully, you know that count was only 1 so you can immediately return tmp.
• if you completely finish the outer loop, that means no values occurred just once, so return -1. This means that -1 should not be a legitimate value. If it is, some other method of showing this should be used.
• finally, if more than one value occurs once, only the first will be returned.

public class SingleNum {
    public static void main(String[] args) {
        int array[] = { 3,3, 2, 2, 4, 5,5 };
        System.out.println(singleNumber(array)); // prints 4
    }
    
    public static int singleNumber(int[] nums) {
        int tmp = 0;
        outer: for (int num : nums) {
            int count = 0;
            tmp = num;
            for (int val : nums) {
                if (tmp == val) {
                    count++;
                    // as soon as count > 1, stop this inner loop
                    if (count > 1) {
                        continue outer;
                    }
                }
            }
            return tmp;
        }
        return -1; // -1 shows no single values.
    }
}