2015-08-11 [pea pod]--Development--1 Noodle

Time: 2015-08-11 15:00 ~ 16:00

Location: Hung Tong Building, posts and telecommunications

Category: Electric surface

1. Ask your CV.

2. Algorithm questions:

Platform: Collabedit

/*

1. Converting a string to an integer
*/

#include <string>#include<climits>usingSTD::string;stringTrimConst string&str) {    intFirst_index =0;  while(First_index <str.size ()) {        if(Str[first_index]! =' ')             Break; First_index++; }    intLast_index = Str.size ()-1;  while(Last_index >=0) {        if(Str[last_index]! =' ')             Break; Last_index--; }    returnStr.substr (First_index, Last_index-first_index +1);}//1, "abc", not string//123456789012345, 2, too longintLast_err_code;//"123456", 123456//" -123456"- -123456//" ", 0 last_err_code = 1//"abc", 0 last_err_code = 1//"+" 0 last_err_code = 1//-0 Last_err_code = 1//"123456789012345", 0 last_err_code = 2intString_to_int (Const string&str) {    intindex =0; BOOLIs_negative =false; Long LongRET =0; Const stringNum_str =trim (str); if(num_str.size () = =0) {Last_err_code=1; return 0; }    if(num_str[0] =='-') {is_negative=false; Index=1; } Else if(num_str[0] =='+') {is_negative=true; Index=1; }    if(Index = =1&& index = =num_str.size ()) {        // "+"        // "-"Last_err_code =1; return 0; }     while(Index <num_str.size ()) {        if(Num_str[index] <'0'|| Num_str[index] >'9') {            //"ABC"//"123CBD"Last_err_code =1; return 0; } ret= RET *Ten+ (int) (Num_str[index]-'0'); if(Ret > (Long Long) {Int_max) {Last_err_code=2; return 0; } Index++; }        returnIs_negative? -1* (int) RET: (int) ret;}

3. Algorithm questions:

Platform: Collabedit

/*
Give two increments of integers a and B with lengths of N and M, respectively. Find out the small number of K in both arrays.
For example: A = [4, 8, ten], B = [2, 3, 4, 5], K = 4. The number of the first K small is 3.
*/

intLast_err_code;//Return-1 when error and set Last_err_codeintFind_kth_small (intA[],intNintB[],intMintK) {    intAns =0; intIndex_a =0; intIndex_b =0; intindex =1; if(N <0|| M <0|| K > n + M | | K <=0) {Last_err_code=1; return-1; }         while(Index <K) {if(Index_a < n && Index_b <m) {if(A[index_a] <B[index_b]) index_a++; ElseIndex_b++; } Else if(Index_a <N) {index_a++; } Else{Index_b++; } Index++;//ERROR forgot to increase the index at the time!     }    if(Index_a < n && Index_b <m) ans= A[index_a] < B[index_b]?A[index_a]: B[index_b]; Else if(Index_a <N) Ans=A[index_a]; Elseans=B[index_b]; returnans;}

Time complexity: O (K)

The interviewer lets optimize:

intLast_err_code;//Return-1 when error and set Last_err_codeintFind_kth_small (intA[],intNintB[],intMintK) {    intA_left =0; intB_left =0; intk_2; inta_k_2; intb_k_2; if(N <0|| M <0|| K > n + M | | K <=0) {Last_err_code=1; return-1; }     while(K >2) {k_2= (K-1) /2; A_k_2= A_left +k_2; A_k_2= A_k_2 < n-1? A_k_2:n-1; B_k_2= B_left +k_2; B_k_2= B_k_2 < M-1? B_k_2:m-1; if(A[a_k_2] = =B[b_k_2]) {            returnA[a_k_2]; } Else if(A[a_k_2] <B[b_k_2]) {K-= A_k_2-A_left; A_left=a_k_2; if(A_left = = N-1)                returnB[b_left + K-2]; } Else{K-= B_k_2-B_left; B_left=b_k_2; if(B_left = = M-1)                returnA[a_left + K-2]; }    }    if(K = =1)        returnA[a_left] < B[b_left]?A[a_left]: B[b_left]; returnA[a_left] > B[b_left]?A[a_left]: B[b_left];}

2015-08-11 [pea pod]--Development--1 Noodle