Implement strStr

Implement strStr()

Implement strStr().

Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

这里用的BF算法实现的，KMP待写...

public class Solution {
    public int strStr(String haystack, String needle) {
        boolean found = true;
        int index = -1;
        if(0 == haystack.length() && 0 == needle.length())
            return 0;
        if(0 == haystack.length() && 0 == needle.length())
            return index;
        
        for(int i = 0; i <= haystack.length() - needle.length(); i++){
            int k = i;
            found = true;
            for(int j = 0; j < needle.length(); j++){
                if(haystack.charAt(k) == needle.charAt(j)){
                    k++;
                    continue;
                }
                else{
                    j = 0;
                    found = false;
                    break;
                }
            }//for
            if(found){
                index = i;
                break;
            }
        }
        
        return index;
    }
}

 今天看了一下KMP然后用KMP写了一个，思路还算清晰就是代码有点长

public class Solution {
    public int strStr(String haystack, String needle) {
        int i = 0;
        int j = 0;
        int next[] = new int[needle.length()];
        if(haystack.length() == 0 && 0 == needle.length())
            return 0;
        if(needle.length() == 0)
            return 0;
        if(0 == needle.length())
            return - 1;
        getNext(needle, next);
        //showNext(next);
        while(i < haystack.length() && j < needle.length()){
            if(j == -1 || haystack.charAt(i) == needle.charAt(j)){
                i++;
                j++;
            }
            else{
                j = next[j];
            }
        }
        if(j == needle.length())
            return i - needle.length();
        return  -1;
    }
        
    /**
     * 计算next[]的值
     * @param pattern
     * @param next
     */
    public void getNext(String pattern, int next[]){
        int k = -1;
        int j = 0;
        next[0] = -1;
        
        while(j < pattern.length() - 1){
            if(-1 == k || pattern.charAt(j) == pattern.charAt(k)){
                k++;
                j++;
                if(pattern.charAt(j) == pattern.charAt(k)){//如果两个相等，递推一下，少一次比较，优化后的代码
                    next[j] = next[k];
                }
                else{
                    next[j] = k;
                }
            }
            else{
                k = next[k];
            }
        }
    }
}