[leetcode] Implement strStr()

Implement strStr().

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

https://oj.leetcode.com/problems/implement-strstr/

思路：目测有BF法，BMP法，BM法等。

BF法：已删除。

第二遍记录：

BF法：

public class Solution {
    public String strStr(String hayStack, String needle) {
        if (hayStack == null || needle == null)
            return null;
        char[] t = hayStack.toCharArray();
        char[] p = needle.toCharArray();
        int i=0,j=0;
        while(i<t.length&&j<p.length){
            if(t[i]==p[j]){
                i++;
                j++;
            }else{
                i=i-j+1;
                j=0;
            }
        }
        if(j==p.length){
            return hayStack.substring(i-j);
        }else{
            return null;
        }
        
        
    }
}

KMP法：

　　注意空字符串“”，生成next数组时越界。next[0]=-1;

　　注意相比BF方法，主函数需要判断j=-1的情况，因为next[j]可能等于-1而导致越界。

public class Solution {
    public String strStr(String hayStack, String needle) {
        if (hayStack == null || needle == null)
            return null;
        char[] t = hayStack.toCharArray();
        char[] p = needle.toCharArray();
        int[] next = geneNext(needle);
        int i=0,j=0;
        while(i<t.length&&j<p.length){
            if(j==-1||t[i]==p[j]){
                i++;
                j++;
            }else{
                j=next[j];
            }
        }
        if(j==p.length){
            return hayStack.substring(i-j);
        }else{
            return null;
        }
        
    }
    
    private int[] geneNext(String ps){
        char[] p = ps.toCharArray();
        int[] next = new int[p.length];
        if(next.length==0)
            return next;
        
        next[0]=-1;
        int j=0;
        int k=-1;
        
        while(j<p.length-1){
            if(k==-1||p[j]==p[k]){
                j++;
                k++;
                next[j]=k;
            }else{
                k=next[k];
            }
            
        }
        
        return next;
        
    }
    
    
}

第三遍记录：

　　p为“”直接返回t。

　　注意next数组的求法。

第四遍记录：

BF法都记不清了，给跪了，双指针法。i和j分别指向两个字符串的当前比较字符。

public class Solution {
    public String strStr(String hayStack, String needle) {
        if(hayStack == null || needle == null)
            return null;
        // It will be convenient to change String to char array first
        char[] t = hayStack.toCharArray();
        char[] p = needle.toCharArray();
        
        int i =0,j=0;
        while(i<t.length&&j<p.length){
            if(t[i]==p[j]){
                i++;
                j++;
            }else{
                //don't change j first because i will use j...
                //i-j stands for the head of this comparison, +1 means to start from next char.
                i = i-j+1;
                j=0;
            }
        }
        if(j==p.length)
            return hayStack.substring(i-j);
        else
            return null;
        
    }
}

KMP

基本思想：当失配时，i指针不回溯，只回溯j指针，j指针回溯到什么位置需要根据p字符串的结构计算出。

public class Solution {
    public String strStr(String hayStack, String needle) {
        if(hayStack == null || needle == null)
            return null;
        //if needle is ""
        if(needle.isEmpty())
            return hayStack;
        // It will be convenient to change String to char array first
        char[] t = hayStack.toCharArray();
        char[] p = needle.toCharArray();
        
        int[] next = getNext(needle);
        
        int i =0,j=0;
        while(i<t.length&&j<p.length){
            //don't forget j == -1
            if(j==-1||t[i]==p[j]){
                i++;
                j++;
            }else{
                j=next[j];
            }
        }
        if(j==p.length)
            return hayStack.substring(i-j);
        else
            return null;
        
    }
    
    private int[] getNext(String ps){
        // we assume ps is not ""
        char[] p = ps.toCharArray();
        int[] next = new int[p.length];
        
        next[0]=-1;
        int k =-1;
        int j = 0;
        //be careful p.length-1
        while(j<p.length-1){
            if(k==-1||p[k]==p[j]){
                k++;
                j++;
                next[j] = k;
            }else{
                k=next[k];
            }
            
        }
        
        return next;
    }
    
}

参考：

http://www.cnblogs.com/yjiyjige/p/3263858.html

http://blog.csdn.net/kenden23/article/details/17029625

http://fisherlei.blogspot.com/2012/12/leetcode-implement-strstr.html