今天在网上看到一个好玩的题目哪来练练哈

最新推荐文章于 2025-05-18 23:13:05 发布

ansj

最新推荐文章于 2025-05-18 23:13:05 发布

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分类专栏：算法讨论文章标签： J#

本文链接：https://blog.youkuaiyun.com/ansjsun/article/details/83534976

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本文介绍了一种用于找出两个字符串中连续相同字符的算法，并通过实例演示如何实现这些匹配段的查找及排序过程。该算法适用于文本处理领域，尤其在字符串相似度比较等场景中有较好的应用。

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[quote]
好像描述的由点问题，再细化点吧！
1、找出连续相同的字符个数最少为min的字符串
2、对字符串排序，这里的排序不是指按字母排序，而是按“连续相同的字符个数”排序
比如：str1="abcdbcebcgh" str2="eabcfbcxbcv" min=2
所得的结果希望是：
NO1. str1: start 0; end 2; str2: start 1; end 3
the same string: abc
NO2. str1: start 4; end 5; str2: start 5; end 6
the same string: bc
NO3. str1: start 4; end 5; str2: start 8; end 9
the same string: bc
NO4. str1: start 7; end 8; str2: start 5; end 6
the same string: bc
NO5. str1: start 7; end 8; str2: start 8; end 9
the same string: bc
因为abc比bc长，所以不在abc查找bc了，而其余的bc因为都相同，所以都需要输出

[/quote]

去重没写..



public class Test {
	public static void main(String[] args) {
		result("abcdbcebcgh","eabcfbcxbcv",2) ;
		Iterator it = set.iterator() ;
		int i= 0 ;
		while(it.hasNext()){
			i++ ;
			System.out.println("NO"+i+". "+it.next());
		}
	}

	private static char[] chars2 = null ;
	private static Set<Str> set = new TreeSet<Str>()  ;
	public static void result(String str1 , String str2 , int min){
		chars2 = str2.toCharArray() ;
		char[] chars1 = str1.toCharArray() ;
		for (int j = 0; j < chars1.length-min+1; j++) {
			for (int i = min; i < chars1.length-j+1; i++) {
				find(new String(chars1,j,i),j,i+j-1) ;
			}
		}

	}
	public static void find(String str,int begin ,int end){
		char[] cs = str.toCharArray() ;
		int k = 0 ;
		for (int i = 0; i < chars2.length; i++) {
			if(k==cs.length){
				set.add(new Str(str,begin,(i-k),end,(i-1))) ;
				k=0 ;
			}
			if(chars2[i]==cs[k]){
				k++ ;
			}else{
				k=0 ;
			}
		}
	}
}

class Str implements Comparable<Str>{
	private String str ;
	private int begin1 ;
	private int begin2 ;
	private int end1 ;
	public Str(String str, int begin1, int begin2, int end1, int end2) {
		this.str = str;
		this.begin1 = begin1;
		this.begin2 = begin2;
		this.end1 = end1;
		this.end2 = end2;
	}
	private int end2 ;
	public String getStr() {
		return str;
	}
	public void setStr(String str) {
		this.str = str;
	}
	public int getBegin1() {
		return begin1;
	}
	public void setBegin1(int begin1) {
		this.begin1 = begin1;
	}
	public int getBegin2() {
		return begin2;
	}
	public void setBegin2(int begin2) {
		this.begin2 = begin2;
	}
	public int getEnd1() {
		return end1;
	}
	public void setEnd1(int end1) {
		this.end1 = end1;
	}
	public int getEnd2() {
		return end2;
	}
	public void setEnd2(int end2) {
		this.end2 = end2;
	}
	public int compareTo(Str o) {
		// TODO Auto-generated method stub
		if(this.str.length()>o.getStr().length()){
			return -1 ;
		}else{
			return 1 ;
		}
	}
	public String toString(){
		return "str1: start "+begin1+"; end "+end1+"; str2: start "+begin2+"; end "+end2+"  \n  the same string:"+str;
	}
}