剑指offer题目练习一

来源：互联网发布：手机oa软件编辑：程序博客网时间：2024/06/01 08:52

看见了一道二维数组找数的题，已排好序的数组（从左至右从上到下，都是由小变大的）让找数，瞬间就出思路了，并没有必要去看他的解释，两次二分就搞定了。

#include<cstdio>#include<iostream>using namespace std;void sreach(int num[][100], int row, int line, int goal){    int i=0,j=row-1,mid;    while((j-i)>1)    {        mid=(i+j)/2;        if(num[mid][0]>goal) j=mid;        else i=mid;    }    int a,b;    a = goal>=num[j][0] ? j : i;    i=0,j=line-1;    while((j-i)>1)    {        mid=(i+j)/2;        if(num[a][mid]>goal) j=mid;        else i=mid;    }

if(num[a][b]!=goal){
cout<<"not found!"<<endl;
}

else cout<<a<<","<<b<<endl;

}int main(){    int n,m;    int number[100][100];    while(scanf("%d%d",&m,&n))    {        for(int i=0;i<n;i++)            for(int j=0;j<m;j++)            scanf("%d",&number[i][j]);        int x;        cin>>x;        sreach(number, n, m, x);    }    return 0;}/*6 61 2 3 4 5 610 12 15 16 18 2122 23 26 27 29 3032 33 34 35 36 3740 41 43 44 47 5050 51 55 56 57 585 51 2 3 4 510 12 15 16 1822 23 26 27 2932 33 34 35 3640 41 43 44 47*/

还有一道插入字符串，把空格都换成%D%，直接链表搞定，不让用链表就用JAVA的linkedlist写。一个函数就够了。

#include<cstdio>#include<iostream>using namespace std;struct node{    char data;    node *next;}*root;   // 指针没有盘空void setuplist(char s[]){    node *temp,*n;    n=root;    int i=0;    while(s[i]!='\0')    {        temp = new node;        temp->next=NULL;        temp->data=s[i];        n->next=temp;        n=temp;        i++;    }}void insertwords(){    node *n;    node *temp1,*temp2;    n=root->next;    while(n->next)    {        if(n->data==' ')        {            n->data='%';            temp1 = new node;            temp1->data = 'd';            temp2 = new node;            temp2->data = '%';            temp1->next=temp2;            temp2->next=n->next;            n->next=temp1;        }        else n=n->next;    }}void del(node *n){    if(n->next) del(n->next);    delete n;}int main(){    char s[100];    while(gets(s))    {        root = new node;        root->next = NULL;        setuplist(s);        insertwords();        node *t=root->next;        while(t->next)        {            cout<<t->data;            t=t->next;        }        cout<<endl;        del(root);    }}

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