"Sword Point offer" topic 38 number of occurrences in a sorted array

Ideas:

Should be used to find the first and last occurrence of the number of two points, subtract. O (LOGN)

intFindleft (intA[],intNintnum) {    intL =0, R = N-1;  while(L <=r) {intm = L + (r-l)/2; if(a[m] = = num)//The difference between the common binary search equals here        {            if(M = =0|| A[m-1]! = num)//If this is the first number or the number in front of it is not Num then this position is where the first num appears.                returnm; Else   //otherwise num in the left halfr = M-1; }        Else if(A[m] <num) {L= m +1; }        Else{R= M-1; }    }    return-1;//didn't find}intFindright (intA[],intNintnum) {    intL =0, R = N-1;  while(L <=r) {intm = L + (r-l)/2; if(A[m] = =num) {            if(M = = N-1|| A[m +1]! = num)////If this is the last number or the number behind it is not Num then this position is where the last num appears.                returnm; Else //otherwise the last num in the right halfL = m +1; }        Else if(A[m] <num) {L= m +1; }        Else{R= M-1; }    }    return-1;//didn't find}intTimesintA[],intNintnum) {    if(A = = NULL)return 0; intL =Findleft (A, n, num); intR =Findright (A, n, num); return(L = =-1|| r = =-1) ?0: R-l +1;}intMain () {inta[Ten] = {1,2,3,3,3,3,4,5}; intAns = times (A,8,5); return 0;}

The first thing to write is to use normal binary search to find a specified number and then expand Around, O (N) is not good.

//O (N) solution is not goodintAppeartimes (intA[],intNintnum) {    intL =0; intr = N-1; intpos =-1;//The subscript for the found Num     while(L <= R)//Two-point lookup to find a num    {        intm = L + (r-l)/2; if(A[m] = =num) {POS=m;  Break; }        Else if(A[m] <num) L= m +1; ElseR= M-1; }    if(pos = =-1)//no such number        return 0; L= R =POS;  while(L >=0&& a[l] = = num)//find the left borderl--;  while(R < n && a[r] = = num)//find the right borderr++; returnR-l-1;}

"Sword Point offer" topic 38 number of occurrences in a sorted array