jq字符串中查找最长并且出现两次以上的一连串字符

[qq_38892057]

例如:abcd12345bnfdvvabcd12345hhnvgabcd1234567abc 查找字符串挨着的字符串 出现两次的以上的最长字符串优先满足条件 上面查找结果为abcd12345 出现次数为3次

[十八道胡同]

import java.util.HashMap;
import java.util.Map;

public class Test04 {
    public static void main(String[] args) {
        String ss = "abcd12345bnfdvvabcd12345hhnvgabcd1234567abc";
        new Test04().findSubString(ss, 1, 2);
    }

    /*
     * minLength 字符串最短长度
     * minCount 最少出现的次数
     * */
    void findSubString(String str, int minLength, int minCount) {
        Map<Integer, String> result = new HashMap<Integer, String>();
        for (int i = 0; i < str.length(); i++) {
            for (int j = i + minLength; j < str.length(); j++) {
                String tmpSur = str.substring(i, j);
                int count = findCount(str, tmpSur);
                if (count >= minCount) {
                    if (!result.containsKey(count)) {
                        result.put(count, tmpSur);
                    } else {
                        if (tmpSur.length() > result.get(count).length()) {
                            result.remove(count);
                            result.put(count, tmpSur);
                        }
                    }
                }
            }
        }
        System.out.println(result);
        Integer finalCount = 0;
        String finalStr = "";
        for (Map.Entry<Integer, String> entry : result.entrySet()) {
            if (entry.getValue().length() > finalStr.length()) {
                finalStr = entry.getValue();
                finalCount = entry.getKey();
            }
        }
        System.out.println("finalStr=" + finalStr + ",出现了" + finalCount + "次");

    }

    public int findCount(String src, String des) {
        int index = 0;
        int count = 0;
        while ((index = src.indexOf(des, index)) != -1) {
            count++;
            index++;
        }
        return count;
    }
}

[qq593467574]

function test(){
		var str="abcd12345bnfdvvabcd12345hhnvgabcd1234567abc";
		var count=3;
		alert(getSub(str,count));
	
	}
	function getSub(test,count){
	  var length=test.length; //要计算的字符串总长度
	  var startLen=parseInt(test.length/count); //要查找的字符串最大长度
	  for(var len=startLen;len>0;len--){
		for(var index=0;index<=length-len;index++){
			//获取子字符串
			var subStr=test.substring(index,index+len);
			console.log(subStr+",len="+len+"|index="+index);
			var replaceStr=test.replace(new RegExp(subStr,'g'),"");
			if((length-replaceStr.length)/len>=count){
				return subStr;
			}
		}
	}
	return "No val";
	}

[qybao]

给你个思路吧 遍历所有的子串，判断子串出现的次数，如果出现2次以上，并且当前子串长度大于之前的子串，则把当前的子串替换之前的子串 最后得到的子串就是结果 我这里偷懒了，用Map保存所有出现2次以上的子串，最后排序选出最长的子串。 LZ可根据我上面的思路自己再改改吧 (注释部分的代码就是思路的代码)

import java.util.*;
import java.util.regex.*;
public class Test {
	public static void test4() {
		String s = "abcd12345bnfdvvabcd12345hhnvgabcd1234567abc";
		Map<String, Integer> map = new HashMap<String, Integer>();
                //String result = "";
                //int times = 0;
		for (int i=0; i<s.length(); i++) {
			for (int j=i+1; j<=s.length(); j++) {
				String ss = s.substring(i,j);
				if (map.containsKey(ss)) {
					continue;
				}
				Pattern p = Pattern.compile(ss);
				Matcher m = p.matcher(s);
				int count = 0;
				while (m.find()) {
					count++;
				}
				if (count >= 2) {
					map.put(ss, count);
                                        //if (ss.length() > result.length()) {
                                        //    result = ss;
                                        //    times = count;
                                        //}
				}	
			}
		}
		//System.out.printf("max leng substring: %s, appear %d times.\n", result, times);
		if (map.size() > 0) {
			List<String> list = new ArrayList<String>(map.keySet());
			list.sort(new Comparator<String>() {
				public int compare(String s1, String s2) {
					return (s1.length() < s2.length()) ? 1 : -1;
				}
			});
			String key = list.get(0);
			System.out.printf("max leng substring: %s, appear %d times.\n", key, map.get(key));
		} else {
			System.out.printf("not found.\n");
		}
	}
}

[qq_38892057]

大神 能不能帮再改动一下