Java面试题：寻找coder；

题目描述

请设计一个高效算法，再给定的字符串数组中，找到包含"Coder"的字符串(不区分大小写)，并将其作为一个新的数组返回。结果字符串的顺序按照"Coder"出现的次数递减排列，若两个串中"Coder"出现的次数相同，则保持他们在原数组中的位置关系。

给定一个字符串数组A和它的大小n，请返回结果数组。保证原数组大小小于等于300,其中每个串的长度小于等于200。同时保证一定存在包含coder的字符串。

测试样例：

["i am a coder","Coder Coder","Code"],3

返回：["Coder Coder","i am a coder"]

public class BinarySearch {public static void main(String[] args) {BinarySearch b=new BinarySearch();//测试用例://["coder","dccoderrlcoderxxpicoderhcoderbiwcoderdcoderrcodermcoderdbvcodertrwvycoderimvcoderuswfccoderczecoderczncoderkfuehcoderocoderiuvccoderfwcodervdiycoderifqjcoder","vxroicoderdqcoderfvcodermtyrcoderlcoderwrygcoder","hcoderwzmjccoderamfmvcoderazmcoderhcodersnuccoderceocodermsmifcoderpwpcodertqbqcoderentbcoderxsgpkcoderrqrbcoderucoder"],4//对应输出应该为://["dccoderrlcoderxxpicoderhcoderbiwcoderdcoderrcodermcoderdbvcodertrwvycoderimvcoderuswfccoderczecoderczncoderkfuehcoderocoderiuvccoderfwcodervdiycoderifqjcoder","hcoderwzmjccoderamfmvcoderazmcoderhcodersnuccoderceocodermsmifcoderpwpcodertqbqcoderentbcoderxsgpkcoderrqrbcoderucoder","vxroicoderdqcoderfvcodermtyrcoderlcoderwrygcoder","coder"]int num=4;String[] strA={"coder","codercodercoder","codercodercoder55","codercodercoder66"};//String[] strA={"coder","codercodercoder","dccoderrlcoderxxpicoderhcoderbiwcoderdcoderrcodermcoderdbvcodertrwvycoderimvcoderuswfccoderczecoderczncoderkfuehcoderocoderiuvccoderfwcodervdiycoderifqjcoder","vxroicoderdqcoderfvcodermtyrcoderlcoderwrygcoder","hcoderwzmjccoderamfmvcoderazmcoderhcodersnuccoderceocodermsmifcoderpwpcodertqbqcoderentbcoderxsgpkcoderrqrbcoderucoder"};//String[] strA={"dccoderrlcoderxxpicoderhcoderbiwcoderdcoderrcodermcoderdbvcodertrwvycoderimvcoderuswfccoderczecoderczncoderkfuehcoderocoderiuvccoderfwcodervdiycoderifqjcoder","hcoderwzmjccoderamfmvcoderazmcoderhcodersnuccoderceocodermsmifcoderpwpcodertqbqcoderentbcoderxsgpkcoderrqrbcoderucoder","vxroicoderdqcoderfvcodermtyrcoderlcoderwrygcoder","coder"};String[] strB=b.findCoder(strA, num);for(int i=0;i<strB.length;i++){System.out.println(strB[i]);}}    public String[] findCoder(String[] A, int n) {    String[] BB=new String[n];    int k;    int m=0;        for(k=0;k<n;k++){    if(A[k].contains("coder") || A[k].contains("Coder") ){    int count=0;    int step=0;    while(step<A[k].length()-4){    if(A[k].substring(step, step+5).equalsIgnoreCase("coder")){    count++;    step=step+5;    }else{    step++;    }    }    BB[m]=A[k]+" "+count;//将包含coder的字符串放入BB字符串数组中；    m++;    }//end if    }//end for    //将非空字符串排序，排序方法是选择排序    String tmp=null;    String[] strB=new String[m];    for(int i=0;i<m;i++){//需要稳定排序    for(int j=m-1;j>i;j--){        String[] bb=BB[j].split(" ");        String[] cc=BB[j-1].split(" ");        int t=Integer.parseInt(bb[bb.length-1]);int t1=Integer.parseInt(cc[cc.length-1]);    if(t>t1){    tmp=BB[j];    BB[j]=BB[j-1];    BB[j-1]=tmp;    }    }    }    for(int i=0,yy=0;i<m;i++,yy++){    String[] dd=BB[i].split(" ");    int countactLong=dd[dd.length-1].length()+1;strB[yy]=BB[i].substring(0,BB[i].length()-countactLong);    }return strB;    }}