[Leetcode OJ] Single number one of the--given an array of integers, every element appears twice except in. Find the single one.

1 classSolution {2  Public:3     intSinglenumber (intA[],intN) {4         inti,j;5          for(i=0; i<n; i++)6         {7              for(j=i+1; j<n; J + +)8             {9                 if(a[j]==A[i])Ten                 { One                     inttemp = a[i+1]; Aa[i+1] =A[j]; -A[J] =temp; -i++; the                      Break; -                 } -             } -             if(j==N) +                 returnA[i];  -         } +     } A};

The above is the traditional method, the time complexity is O (N2), the space complexity is O (1).

1 classSolution {2  Public:3     intSinglenumber (intA[],intN) {4         intresult=0;5          for(intI=0; i<n; i++)6result = result ^A[i];7         returnresult;8     }9};

The time complexity and space complexity are all O (1) by using bitwise XOR OR operation.

The nature of the commutative law and the binding law is used for the XOR operation.

Exchange Law: a^b = B^a

Binding law: (a^b) ^c = a^ (b^c)

In addition, a^a=0.