1. 找出数组中的单身狗OddOccurrencesInArray Find value that occurs in odd number of elements....

最新推荐文章于 2022-04-09 21:50:55 发布

weixin_34318326

最新推荐文章于 2022-04-09 21:50:55 发布

阅读量224

点赞数

文章标签： java

本文介绍了一种在数组中找到唯一未配对元素的方法。该方法适用于数组中大部分元素成对出现的情况，仅有一个元素单独存在。通过排序数组并检查相邻元素来实现，最终返回未配对的元素。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

找出数组中的单身狗；

1. OddOccurrencesInArray
Find value that occurs in odd number of elements.

A non-empty zero-indexed array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.

For example, in array A such that:

  A[0] = 9  A[1] = 3  A[2] = 9
  A[3] = 3  A[4] = 9  A[5] = 7
  A[6] = 9
the elements at indexes 0 and 2 have value 9,
the elements at indexes 1 and 3 have value 3,
the elements at indexes 4 and 6 have value 9,
the element at index 5 has value 7 and is unpaired.
Write a function:

class Solution { public int solution(int[] A); }

that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.

For example, given array A such that:

  A[0] = 9  A[1] = 3  A[2] = 9
  A[3] = 3  A[4] = 9  A[5] = 7
  A[6] = 9
the function should return 7, as explained in the example above.

Assume that:

N is an odd integer within the range [1..1,000,000];
each element of array A is an integer within the range [1..1,000,000,000];
all but one of the values in A occur an even number of times.
Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.

package com.code;

import java.util.Arrays;

public class Test02 {
    
    public int solution(int[] A) {
        // write your code in Java SE 8
        int size = A.length;
        if(size%2==0){
            return -1;
        }
        //System.out.println(Arrays.toString(A));
        Arrays.sort(A);
        //System.out.println(Arrays.toString(A));
        if(size==1){
            return A[0];
        }
        if(A[0]!=A[1]){
            return A[0];
        }
        if(A[size-1] != A[size-2]){
            return A[size-1];
        }
        for(int i=2;i<size-1;i=i+2){
            if(A[i]!=A[i+1]){
                return A[i];
            }
        }
        return 0;
    }
    
    public static void main(String[] args) {
        Test02 t02 = new Test02();
        int[] a = {9,3,9,3,5};
        
        System.out.println(t02.solution(a));
        int [] b = {1,1,2,2,3,4,4};
        
        System.out.println(t02.solution(b));
        
    }
}