算法面试题解答（四）

1. 求两个字符串的最大公共字符串

#include <iostream>

using namespace std;

void getLength(const char *src1, const char *src2, int *l1, int *l2)
{
    if(src1 == NULL || src2 == NULL || strlen(src1) == 0 || strlen(src2) == 0)
    {
        *l1 = 0;
        *l2 = 0;
        return ;
    }
    int pl1 = 0,pl2 = 0;
    
    int tpl1 = 0,tpl2 = 0;
    int tpl3 = 0,tpl4 = 0;
    if(src1[0] == src2[0])
    {
        getLength(src1+1,src2+1, &tpl1, &tpl2);
        pl1 = tpl1+1;
    }
    else
    {
        pl1 = 0;
        getLength(src1+1,src2, &tpl1, &tpl2);
        getLength(src1,src2+1, &tpl3, &tpl4);
    }
    *l1 = pl1;
    *l2 = tpl2>tpl4?tpl2:tpl4;
    if(*l2<pl1)
        *l2 = pl1;
}

int main(int argc, char **argv)
{
    int l1 =0 , l2  =0 ;
    getLength("abcdefg","cdefbcderg", &l1, &l2);
    cout<<l2<<endl;
}

 

2. Convert a binary search tree to a circular sorted linked list. The highest valued node should point to the lowest valued node at each step. 下面只是把BST转为单向链表，而且没有处理left指针，可以再完善一下。

#include <iostream>
using namespace std;
struct BSTNode 
{
    BSTNode(int v){value = v;left = right = NULL;}
    int value;
    BSTNode *left;
    BSTNode *right;
};

void ConvertBST2CiruluarLinkedList(BSTNode *root, BSTNode **head, BSTNode **tail)
{
    if(root == NULL)
        return;
    BSTNode *headl = NULL, *taill = NULL;
    if(root->left != NULL)
    {
        ConvertBST2CiruluarLinkedList(root->left,&headl, &taill);        
    }
    if(headl != NULL)
    {
        *head = headl;
        taill->right = root;
    }
    else
    {
        *head = root;
    }
    BSTNode *headr = NULL, *tailr = NULL;
    if(root->right !=NULL)
    {
        ConvertBST2CiruluarLinkedList(root->right,&headr, &tailr);
    }
    if(headr != NULL)
    {
        root->right = headr; //anyway, we should use root to make connection with right linked list;
        tailr->right = *head;
        *tail = tailr;
    }
    else
    {
        *tail = root;
        (*tail)->right = *head;
    }
}

int main(int argc, char **argv)
{
    BSTNode *a4 = new BSTNode (4);
    BSTNode *a3 = new BSTNode (3);
    BSTNode *a2 = new BSTNode (2);
    BSTNode *a1 = new BSTNode (1);
    BSTNode *a5 = new BSTNode (5);
    a2->left = a1;
    a2->right = a3;
    a4->left = a2;
    a4->right = a5;
    BSTNode * head = NULL, *tail = NULL;
    ConvertBST2CiruluarLinkedList(a4,&head,&tail);
}