hulu面试题2

一、填空

侧重逻辑思维，没有语言、具体技术考察，大部分属于组合数学、算法。比较基本的知识点有二元树节点树、最小生成树、Hash函数常用方法等。

二、编程题

1、正整数剖分

2、AOE关键路径

3、二元树前序、中序求后序

4、大整数加

 

//正整数剖分#include <stdio.h>int func(int n, int k, int max){    int min = (int)((n+k-1)/k);    if(k==1)        return 1;    int count = 0;    for(int i=min;i<max;i++){        count += func(n-i, k-1, max-i);    }    return count;}int main(){    int ans;    int i = 10;    //ans = func(10, 3, 10);    ans = func(4, 3, 4);    printf("%d\n", ans);    return 0;}

 

//AOE#include <stdio.h>#include <stdlib.h>#include <string.h>void AOE(int adj[][9], int n){    int *e = (int*)malloc(sizeof(int)*n);    int *l = (int*)malloc(sizeof(int)*n);    e[0]=0;    for(int i=1;i<n;i++){        int max = 0;        for(int j=0;j<i;j++){            if(adj[j][i]!=0 && adj[j][i]+e[j]>max){                max = adj[j][i] + e[j];            }        }        e[i]=max;    }    l[n-1]=e[n-1];    for(int i=n-2;i>=0;i--){        int min = 999999;        for(int j=n-1;j>i;j--){            if(adj[i][j]!=0 && l[j]-adj[i][j]<min){                min = l[j]-adj[i][j];            }        }        l[i]=min;    }    for(int i=0;i<n;i++){        if(e[i]==l[i])            printf("%d\t", i+1);    }    printf("\n");    free(e);    free(l);}int main(){    const int size = 9;    int adj[size][size];    memset(adj, 0, sizeof(int)*size*size);    adj[0][1] = 6, adj[0][2]=4, adj[0][3]=5;    adj[1][4] = 1, adj[2][4]=1, adj[3][5]=2;    adj[4][6] = 6, adj[4][7]=5, adj[5][7]=4;    adj[6][8] = 2, adj[7][8]=4;    AOE(adj, size);    return 0;}

//二元树前序、中序打印后序#include <stdio.h>#include <cstring>#include <stack>using namespace std;void dumpPost(const char* pre, const char* mid){    int n = strlen(pre);    if(n==1){        printf("%c\t", pre[0]);        return;    }    int i;    for(i=0;i<n;i++){        if(mid[i]==pre[0])            break;    }    char lpre[i], lmid[n-i-1];    char rpre[i], rmid[n-i-1];    memcpy(lpre, pre+1, sizeof(char)*i);    memcpy(lmid, mid, sizeof(char)*i);    memcpy(rpre, pre+i+1, sizeof(char)*(n-i-1));    memcpy(rmid, mid+i+1, sizeof(char)*(n-i-1));    lpre[i] = lmid[i] = '\0';    rpre[n-i-1] = rmid[n-i-1] = '\0';    dumpPost(lpre, lmid);    dumpPost(rpre, rmid);    printf("%c\t", pre[0]);}int main(){    const char* preOrder = "ABDEC";    const char* midOrder = "DBEAC";    const char* postOrder = "DEBCA";    dumpPost(preOrder, midOrder);    printf("\n");    return 0;}

//大整数运算#include <stdio.h>#include <stdlib.h>#include <string.h>void strrev(char* s){    int i=-1;    while(s[++i]!='\0');    for(int j=0;j<i/2;j++){        char tmp = s[j];        s[j] = s[i-j-1];        s[i-j-1]=tmp;    }}void Add(const char*str1, const char* str2, char* ans){    int l1, l2, l;    l1 = strlen(str1);    l2 = strlen(str2);    l = l1>l2 ? l1 : l2;    char* s1 = (char*)malloc(sizeof(char)*(l1+1));    char* s2 = (char*)malloc(sizeof(char)*(l2+1));    memcpy(s1,str1,(l1+1)*sizeof(char));    memcpy(s2,str2,(l2+1)*sizeof(char));    strrev(s1);    strrev(s2);    int i;    int sum, carry;    i=sum=carry=0;    while(i<l){        char a = i<l1?s1[i]:'0';        char b = i<l2?s2[i]:'0';        sum = a-'0'+b-'0' + carry;        ans[i] = sum % 10 + '0';        carry = sum / 10;        i++;    }    if(carry!=0)        ans[i++]=carry+'0';    ans[i]='\0';    strrev(ans);    free(s1);    free(s2);}void Mul(const char*str1, const char* str2, char* ans){    int l1, l2, l;    l1 = strlen(str1);    l2 = strlen(str2);    l = l1 + l2;    ans[0]='\0';    char* s1 = (char*)malloc(sizeof(char)*(l1+1));    char* s2 = (char*)malloc(sizeof(char)*(l2+1));    memcpy(s1,str1,(l1+1)*sizeof(char));    memcpy(s2,str2,(l2+1)*sizeof(char));    strrev(s1);    strrev(s2);    char* tmp = (char*)malloc(sizeof(char)*(l1+2));    int s, carry;    s = carry = 0;    for(int i=0;i<l2;i++){        int j;        for(int j=0;j<i;j++)            tmp[j]='0';        for(j=0;j<l1;j++){            s = (s1[j]-'0')*(s2[i]-'0')+carry;            tmp[i+j]=s%10+'0';            carry=s/10;        }        if(carry!=0)            tmp[i+j++]=carry+'0';        tmp[i+j]='\0';        strrev(ans);        strrev(tmp);        Add(ans,tmp, ans);        strrev(ans);    }    strrev(ans);}int main(){    const char a[] = "12345";    const char b[] = "123";    char c[1024];    Add(a,b,c);    printf("a+b=%s\n", c);    Mul(a,b,c);    printf("a*b=%s\n", c);    return 0;}