Question

因此，我试图用一种方法来解决分而治之的原则和多次递归调用。可以，但是我编写的方法的输出有问题。

该方法的目的是返回数组中所有连续数字对的总和。我95％的人在那儿，但没有得到我期望的输出，并且多年来一直试图将我撞到桌子上，试图找出原因。

该数组是：

int[] array = { 11, 6, 87, 32, 15, 5, 9, 21 };

，方法是：

public int consecutivePairsSum_DivideAndConquer(int start, int end, int[] array) {
    int leftSum;
    int rightSum;
    int middle = (start + end) / 2;
    if (start == middle) {
        return array[middle];
    } else {
        leftSum = array[start] + array[start + 1];
        leftSum += consecutivePairsSum_DivideAndConquer(start, middle, array);
    }
    if (middle == end) {
        return array[end];
    } else {
        rightSum = array[middle] + array[middle+1];
        rightSum += consecutivePairsSum_DivideAndConquer(middle+1, end, array);
    }
    return leftSum + rightSum;
}

这是我的方法调用：

System.out.println(rF.consecutivePairsSum_DivideAndConquer(0, array.length-1, array));

我认为这与拆分数组的方式有关，但是没有大量的实验可以为我提供正确的输出。

预期输出：340

实际输出：330

任何欢迎的建议，这真让我发疯！：p

ps任何有用的链接，也可以找到我在其中可以找到可靠的有关递归的可靠在线教程/好书的链接（如果这在SO的范围之内，因为它不了解如何直接帮助解决编程问题）

Answer 1

以下是算法的概述：

基本情况：如果您的数组少于两个元素，则结果为#include <stdio.h> #include <unistd.h> #include <string.h> #include <stdlib.h> #include <errno.h> #include <time.h> #include <sys/stat.h> #include <sys/socket.h> #include <netinet/in.h> #define BUFFER_SIZE 9999 #define HTTP_METHOD "HTTP/1.1 " #define HTTP__OK "200 OK\r\n" #define HTTP__NOT_FOUND "404 Not Found\r\n" #define SERVER_NAME "Server: ECE435\r\n" /* Default port to listen on */ #define DEFAULT_PORT 8080 //modify port to listen on 8080 int main(int argc, char **argv) { int socket_fd,new_socket_fd; struct sockaddr_in server_addr, client_addr; int port=DEFAULT_PORT; int n; socklen_t client_len; char buffer[BUFFER_SIZE]; printf("Starting server on port %d\n",port); /* Open a socket to listen on */ /* AF_INET means an IPv4 connection */ /* SOCK_STREAM means reliable two-way connection (TCP) */ socket_fd = socket(AF_INET, SOCK_STREAM, 0); if (socket_fd<0) { fprintf(stderr,"Error opening socket! %s\n", strerror(errno)); exit(1); } /* Set up the server address to listen on */ /* The memset stes the address to 0.0.0.0 which means */ /* listen on any interface. */ memset(&server_addr,0,sizeof(struct sockaddr_in)); server_addr.sin_family=AF_INET; /* Convert the port we want to network byte order */ server_addr.sin_port=htons(port); /* Bind to the port */ if (bind(socket_fd, (struct sockaddr *) &server_addr, sizeof(server_addr)) <0) { fprintf(stderr,"Error binding! %s\n", strerror(errno)); fprintf(stderr,"Probably in time wait, have to wait 60s if you ^C to close\n"); exit(1); } /* Tell the server we want to listen on the port */ /* Second argument is backlog, how many pending connections can */ /* build up */ listen(socket_fd,5); wait_for_connection: /* Call accept to create a new file descriptor for an incoming */ /* connection. It takes the oldest one off the queue */ /* We're blocking so it waits here until a connection happens */ client_len=sizeof(client_addr); new_socket_fd = accept(socket_fd, (struct sockaddr *)&client_addr,&client_len); if (new_socket_fd<0) { fprintf(stderr,"Error accepting! %s\n",strerror(errno)); exit(1); } while(1){ /* Someone connected! Let's try to read BUFFER_SIZE-1 bytes */ memset( buffer, 0, BUFFER_SIZE ); n = read( new_socket_fd, buffer, ( BUFFER_SIZE-1 ) ); if (n==0){ fprintf( stderr, "Connection to client lost\n\n" ); break; } else if( n < 0 ){ fprintf(stderr,"Error reading from socket %s\n", strerror(errno)); } /* Print the message we received */ printf("Message received: %s\n" ,buffer); const char *PATTERN1 = "GET /"; //first cut to make on buffer const char *PATTERN2 = " HTTP"; //second cut to make on buffer char *target = NULL; //variable to hold the slice we're taking char *start, *end; //defining variables to hold start and end positions if ( start = strstr( buffer, PATTERN1 ) ){ //code to grab a slice of buffer start += strlen( PATTERN1 ); if ( end = strstr( start, PATTERN2 ) ){ target = ( char * )malloc( end - start + 1 ); memcpy( target, start, end - start ); target[end - start] = '\0'; } } if ( target ) printf( "Client requested: %s\n", target ); //code is working to this point. I can tell what file to get. time_t rawtime; struct tm info; time( &rawtime ); struct tm * timeinfo; char timestamp[100]; time_t now = time(0); struct tm tm = *gmtime(&now); strftime(timestamp, sizeof( timestamp ) , "%a, %d %b %Y %H:%M:%S %Z", &tm); //printf("Time is: [%s]\n", timestamp); struct stat file_info; //define statistics structure for file info stat( target, &file_info ); //initiate file_info as the stat structure for target char send_client[9999]; sprintf( send_client, "HTTP/1.0 %s%s\r\nServer: ECE435\r\nLast-Modified: Fri, 08 Sep 2017 04:31:47 GMT\r\nContent-Length: 85\r\nContent-Type: text/html\r\n\r\n", HTTP__OK, timestamp ); char file_path[256]; //limited to 256 char for now sprintf( file_path, "./%s", target); //this is how you can combine char arrays: puts "./" + 'target' into 'file_path' //int fd; //printf( "%ld\r\n" , file_info.st_size ); //this should print the File Size char source[BUFFER_SIZE + 1]; FILE *fp = fopen( file_path, "r"); if (fp != NULL) { size_t newLen = fread(source, sizeof(char), BUFFER_SIZE, fp); if (newLen == 0) { fputs("Error reading file", stderr); } else { source[newLen] = '\0'; /* Just to be safe. */ } fclose(fp); } strcat( send_client, source); /* Send a response */ printf( "\r\n%s\r\n" , send_client ); //print response before sending n = write( new_socket_fd, send_client , strlen(send_client) ) ; if( n < 0 ){ fprintf( stderr, "Error writing. %s\n", strerror(errno)); } } close(new_socket_fd); printf("Done connection, go back and wait for another\n\n"); goto wait_for_connection; /* Try to avoid TIME_WAIT */ // sleep(1); /* Close the sockets */ close(socket_fd); return 0; }（因为没有对）。

否则：将数组分为两半，计算左，右两半的结果，则整个数组的结果将为0（因为这里唯一丢失的对是位于

在Java中，将是这样的：

<result of left half> + <result of right half> + <last element of left half> + <first element of right half>

应称为

int consPairSum(int[] array, int left, int right) {
    if(right <= left + 1) return 0;

    int mid = (left + right) / 2;
    int leftPairSum = consPairSum(array, left, mid);
    int rightPairSum = consPairSum(array, mid, right);
    return leftPairSum + rightPairSum + array[mid - 1] + array[mid];
}

Answer 2

谁说分而治之需要划分为相等的块，您只需要划分为自身相似的问题。差不多有1班轮

static private int doTheThing(int[] list){
    if (list.length==2)
        return list[0]+list[1];
    return list[0]+list[1]+doTheThing(Arrays.copyOfRange(list,1,list.length));
}

在使用分而治之算法在数组中添加连续的整数对时遇到麻烦

2 个答案: