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HDU1238

简单的求最大子串问题,难点在于反向比较的也要算
 
[java] 
package D0710; 
 
/*
 * 求子串问题
 * 思路:
 *1、 获得每组测试数据中最短的字符串
 *2、用最短的字符串的子串进行正相匹配和反向匹配
 *3、得长度最大的子串即可
 * 
 */ 
import java.util.Scanner; 
 
public class HDU1238 { 
 
    public static void main(String[] args) { 
        Scanner sc = new Scanner(System.in); 
        int t; 
        int n; 
        String[] str;// 保存接收输入的字符串 
        String[] str2;// 保存str的反串 
        StringBuffer sb; 
        String s;// 保存每组测试数据中最短字符串 
        int len;// 最后得到的最大子串长度 
        t = sc.nextInt(); 
        while (t-- > 0) { 
            n = sc.nextInt(); 
            str = new String[n]; 
            str2 = new String[n]; 
            // 接收输入 
            for (int i = 0; i < n; i++) { 
                str[i] = sc.next(); 
                sb = new StringBuffer(str[i]); 
                str2[i] = sb.reverse().toString(); 
            } 
            // 得最短字符串, str2中的最短字符串应当和str中一致 
            s = str[0]; 
            for (int i = 0; i < n; i++) { 
                if (str[i].length() < s.length()) { 
                    s = str[i]; 
                    str2[i] = s; 
                } 
            } 
            if (s.equals(str[0])) { 
                str[0] = str2[0] = s; 
            } 
            // 匹配,得最大子串长度 
            len = getLen(s, str, str2); 
            System.out.println(len); 
        } 
 
    } 
 
    private static int getLen(String s, String[] str, String[] str2) { 
 
        int l = str.length; 
        boolean flag = false; 
        int l2 = s.length(); 
        String sub;// 子串 
        String rsub = "";// 最后的子串 
        int i, j, k;// 循环变量 
        for (i = 0; i < l2; i++) { 
            for (j = i + 1; j <= l2; j++) { 
                sub = s.substring(i, j); 
                // 正向匹配 
                for (k = 0; k < l; k++) { 
                    if (str[k].contains(sub)) { 
                        flag = true; 
                    } else { 
                        flag = false; 
                        break; 
                    } 
                } 
                if (flag) { 
                    if (sub.length() > rsub.length()) 
                        rsub = sub; 
                } 
                // 反向匹配 
                for (k = 0; k < l; k++) { 
                    if ( str2[k].contains(sub)) { 
                        flag = true; 
                    } else { 
                        flag = false; 
                        break; 
                    } 
                } 
   &nb

补充:软件开发 , C语言 ,
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