tinglu的专栏

距离：插入：1删除：1替换：2，（1也行，改下代码变2为1）#include using namespace std;const int N = 100;int d[N][N];char s1[N], s2[N];inline int three_min(int x, int y, int z){ int a = x<y?x:y; return a<

72个人，从第9个人开始报数，报到11的人退出，下一位继续从1开始报数，问最后一个人编号。#includeusing namespace std;const int NUM = 72;struct node{ int value; node* next;};void func(){ int i; node* head = new node;

算法本身不复杂，但是提供了一个跟有序数组相关的一个思路：记录首尾指针，由两边向中间缩进。#include using namespace std;bool findPairWithSum(int* a, int len, int sum, int& n1, int& n2){ if(a==NULL || lena[len-1]+a[len-2]) return false;

比归并排序只多了一句输出。#include using namespace std;void getInversePair(int* a, int* b, int low, int mid, int high){ int l=low, m=mid+1, k=low, j; while(l<=mid && m<=high) { if(a[l]>a[m]) { f

#include using namespace std;#define min(x, y) (x)<(y)?(x):(y) char getFirstSingleChar(char* s){ if(s==NULL || *s==0) return 0; int len = strlen(s); unsigned int h_table[256] = {0}; int i

链表翻转。给出一个链表和一个数k，比如链表1→2→3→4→5→6，k=2，则翻转后2→1→4→3→6→5，若k=3,翻转后3→2→1→6→5→4，若k=4，翻转后4→3→2→1→5→6#include using namespace std;struct node{ int data; node* next;};node* init(int n){ n

#include #include using namespace std;void heap_adjust(int *a, int i, int n){ int l, r, big; while(n>=2*i+1) { big = i; l = 2*i+1; if(a[l]>a[i]) b

#include using namespace std;const int N = 20;char s[N][N];char com1[2*N+1];char com2[2*N+1];int comp(const void *v1, const void *v2){ strcpy(com1, (char*)v1); strcat(com1, (char*)v2); str

#include using namespace std;struct node{ int data; node* next;};void remove(node** head, int value){ if(head == NULL) return ; node** n = head; node* cur; while

#include using namespace std;void reverse(char* s,int left, int right){ while(left<right) { swap(s[left], s[right]); left++; right--; }}void reverse_sentence

#include using namespace std;#define min(x, y) (x)<(y)?(x):(y) int getUglyNumber(int index){ if(index<0) return 1; int* nums = new int[index]; nums[0] = 1; int next_id = 1; int* ugly2

#include using namespace std;void func(int *a, int len, int &b1, int &b2){ int a1, a2; b1 = a1 = 0; b2 = a2 = len-1; int l_min, r_max; l_min = a[a1]; r_max = a[a2]; while(a1<a2) { if(a[

#include using namespace std;void huiwen(char* a){ if(a==NULL) return ; int len = strlen(a); bool table[100][100] = {false}; int max, max_id; int i,j; for(i=0; i<len; i++) table[i][i] =

#include using namespace std;void huiwen(char* a, int& max, int& max_id){ if(a==NULL || *a=='\0') { max = 0; max_id = 0; return ; } int len = strlen(a); int i, k1, k2; max = 1; max_id

#include #include using namespace std; //函数功能 ： 从一个字符串中选m个元素//函数参数 ： pStr为字符串， m为选的元素个数， result为选中的//返回值 ： 无void Combination_m(char *pStr, int m, vector &result){ if(pStr == NULL ||

#include #include using namespace std; struct node{ int data; node* lchild; node* rchild;}; node* buildBTree(int* a, int &i){ if(a==NULL || a[i]==-1)

剑指offer用的是字符串，还要苦逼的字符串比较，看是否进位。直接用整型数组来保存。#include #include using namespace std; //一个int保存几位数const int int_num = 2;//每位数最大值，超过这个要进位（不包括最高位）const int int_max = 99; void func(){

#include using namespace std;void func(int* pre, int* mid, int pre_left, int pre_right, int mid_left, int mid_right){ if(pre_left == pre_right) { cout<<pre[pre_left

关键问题：不在同行，同列，斜线上。即对任意两行，!(x[r1]==x[r] || abs(r1-r)==abs(x[r1]-x[r]))。用回溯法，通过此条件进行减枝。#include using namespace std;void print(int *x, int n){ int i, j; for(i=0; i<n; i++) { for(j=0; j<n; j

主要思想，分别考虑每位（个位，十位，不是二进制的位）为1的情况，特别考虑最高位是1和不是1的情况。使用递归，如 f(253)=f(200)+f(50)+f(3)#include #include using namespace std;int f1(int n){ int i, j, k; int num_1=0; for(i=0; i<=n; i++)

#include using namespace std;struct node{ int data; node* lchild; node* rchild;};bool is_banance_tree(node* r, int& height){ if(r == NULL) { height = 0; retu

#include #include using namespace std; struct node{ int data; node* lchild; node* rchild;};void find_path(node* r, int exceptedSum, vector &path, int& curSu

#include using namespace std;void f(){ int a[] = {-6,-8, -5, -9}; int len = sizeof(a)/sizeof(a[0]); int sum = a[0] ; int d = a[0]; int i; for(i=1; i<len; i++) {

#include using namespace std;struct node{ int data; node* next;};node* reverse_link(node* head){ if(head==NULL || head->next==NULL) return head; node* pre; node

#includeusing namespace std;const int contain_num[] = {5,6};const int num_len = sizeof(contain_num)/sizeof(contain_num[0]);const int MAX = 3000;bool b_has[10] = {false};bool if_contain(int n);

#includeusing namespace std;void func(int n){ int i; for(i=2; i<n; ) { if(n%i == 0) { cout<<i<<"*"; n /= i; } else i

递归与非递归实现#includeusing namespace std;struct node{ int value; node* next;};//非递归node* func(node* head){ if(head==NULL || head->next==NULL) return head; node* p1

#include using namespace std;int k=3;int find_single(int* a, int n){ int i, j; int single = 0; int temp[32]; for(i=0; i<32; i++) temp[i] = 0; for(i=0; i<n; i

#include #include using namespace std;const int N = 20;const int MAX_DIS = 10000;int dis[N]; //v0到i的最短路径长度int pre[N]; //节点i的前驱int dd[N][N]; //节点之间距离int sort_id[N]; //记录dis[n]排序后的下标，关键int n

后缀数组中，需要排序字符串：#include #include using namespace std;int comp(const void* p1, const void* p2){ return strcmp((char*)p1, (char*)p2);}int main(){ char s[5][6] = {"abc", "abcd", "bcde",

#include using namespace std;bool is_swap(char* a, int i, int j){ while(i<j) { if(a[i] == a[j]) return false; i++; } return true;}void print_a(char*

#include #include using namespace std;#define N 100char s[N];char suf[N][N];int max_same = 0;int comp(const void *s1, const void* s2){ return strcmp((char*)s1, (char*)s2);}int ge

注意指向指针的指针的应用。#include #include using namespace std;struct node{ int data; node* next;};node* partion(node** head, node* end){ // coutdata<<" "; // if(end==NULL) //

#include using namespace std;class T{public: T():buf(NULL){} T(char *s) { buf = new char[strlen(s) + 1]; strcpy(buf, s); } T(const T& t) { char* t_

#include using namespace std;int main(){ char a[] = "453453236478236783462786"; char b[] = "98928342347896248326"; int lenA = strlen(a); int lenB = strlen(b); int lenR = lenA+lenB; int* r

#include using namespace std;void set_next(int* next, char *s){ int k, j; int len = strlen(s); next[0] = -1; for(k=1; k<len; k++) { j = k-1; while(1) {

#include using namespace std;int next_norepeat(unsigned int n){ int pow, i, j; i=++n; pow=0; while(i) { pow++; i/=10; } int *a = new int[pow+1]; a[0] = 0; for(i=pow, j=n; i>0; i--, j

#include using namespace std;void reverse_str(char* s, int l, int h){ while(l<h) { swap(s[l++], s[h--]); }}void left_rotate_reverse(char* s, int k, int n){ reverse_str(

#include using namespace std; class D{ public: virtual void d(){cout<<"D::d"<<endl;} // int dd; };class ClassA : public virtual D{public:virtual ~ ClassA(){};virtual void FunctionA(){cout

#include using namespace std;#define N 1000int max_len = 0;int max_right_id = 0;char s[N];int c[N];int pre[N];void print(int id){ if(pre[id] == -1) { cout<<max_len<<": "<<