OddOccurrencesInArray

给定一个非空整数数组,数组中含有奇数个元素且大部分元素成对出现,只有一个元素不成对。编写一个函数找到这个未配对的元素。例如,输入数组 [9, 3, 9, 3, 9, 7, 9],函数应返回 7。" 73838562,5551647,Ubuntu VMware上MySQL远程连接配置,"['VMware', '数据库', 'Ubuntu', 'MySQL', '远程访问']

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Desc:
A non-empty 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.

Write an efficient algorithm for the following assumptions:

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.
Copyright 2009–2021 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

Solution

package solution

// you can also use imports, for example:
// import "fmt"
// import "os"

// you can write to stdout for debugging purposes, e.g.
// fmt.Println("this is a debug message")

func Solution(A []int) int {
    // write your code in Go 1.4
    dicMap :=make(map[int]bool)
    for _,v :=range A {
        if _,ok :=dicMap[v];ok {
            if dicMap[v] {
                dicMap[v] =false
            }else{
                dicMap[v] =true
            }
        }else{
                dicMap[v] =true
        }
    }
    for k,v :=range dicMap{
        if v {
            return k
        }
    }
    return 0
}
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