[LeetCode28] Implement strStr()

最新推荐文章于 2017-07-15 13:14:01 发布

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文章标签：

#string #leetcode #KMP #indexof #match

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本文详细介绍了Java中字符串匹配的两种经典算法——KMP算法与滚动哈希法，并提供了Java代码实现。同时，文章还分享了面试中解决此类问题的高效策略。

Implement strStr().

Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack.

in Java, there is an API function name indexof(), it returns index within this string of the first occurrence of the specifiedsubstring.

java 实现

public String strStr(String haystack, String needle) {
		if(haystack.length()<=0) return null;
        int i = haystack.indexOf(needle);
        if(i==-1) return null;
        else{
        	return haystack.substring(i);
        }
    }

another solution:

public String strStr(String haystack, String needle) {
		if(haystack.length()<=0 && needle.length()>0) return null;
        for(int i=0;i<=haystack.length()-needle.length();i++){
        	boolean flag = true;
        	for(int j=0;j<needle.length();j++){
        		if(haystack.charAt(i+j)!=needle.charAt(j)){
        			flag = false;
        			break;
        		}
        	}
        	if(flag){
    			return haystack.substring(i);
    		}
        }
        return null;
    }

Below two solutions is overkill for interview.

third solution: rolling hash

http://blog.youkuaiyun.com/yanghua_kobe/article/details/8914970

fourth solution: KMP

http://www.ruanyifeng.com/blog/2013/05/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm.html

For c++

strightforward use point scan from begin to end, but this complexity is O(M*N)

char *strStr(char *haystack, char *needle) {
      if(!haystack || !needle) return NULL;
    int hlength = strlen(haystack);
    int nlength = strlen(needle);
    if(hlength<nlength) return NULL;
    for(int i=0; i<hlength-nlength+1;i++){
        char *p = &haystack[i];
        int j=0;
        for(;j<nlength;j++){
            if(*p == needle[j])
                p++;
            else
                break;
        }
        if(j == nlength)
            return &haystack[i];
    }
    return NULL;
    }

the best solution is KMP, complexity is O(N)