LeetCode 28 — Implement strStr()（C++ Java Python）

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题目：http://oj.leetcode.com/problems/implement-strstr/

Implement strStr().

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

题目翻译：

实现strStr()。
返回指向needle在haystack中第一次出现位置的指针，如果needle不是haystack的一部分，则返回null。

分析：
采用暴力方法，注意下标不要越界。（KMP待补充）

C++实现1：

class Solution {public:    char *strStr(char *haystack, char *needle) {if(*needle == NULL) {return haystack;}char *p1;char *p2 = needle + 1; char *p1_advance = haystack;while(*p2){++p1_advance;++p2; }while(*p1_advance) {p1 = haystack;p2 = needle;while(*p2 && *p1 == *p2){++p1;++p2;}if(*p2 == NULL) {return haystack;}++haystack;++p1_advance;}return NULL;    }};

C++实现2：

class Solution {public:    char *strStr(char *haystack, char *needle) {    std::string str(haystack);    std::string target(needle);    if(target.empty())    {    return haystack;    }    int len1 = str.length();    int len2 = target.length();    if(len1 < len2)    {    return NULL;    }    for(int i = 0; i <= len1 - len2; ++i)    {    int j = 0;    int k = i;    while(j < len2 && target[j] == str[k])    {    k++;    j++;    }    if(j == len2)    {    return (char*)str.substr(i).c_str();    }    }    return NULL;    }};

Java实现：

public class Solution {    public String strStr(String haystack, String needle) {if (needle != null && needle.isEmpty()) {return haystack;}int len1 = haystack.length();int len2 = needle.length();if (len1 < len2) {return null;}for (int i = 0; i <= len1 - len2; ++i) {int j = 0;int k = i;while (j < len2 && needle.charAt(j) == haystack.charAt(k)) {j++;k++;}if (j == len2) {return haystack.substring(i);}}return null;    }}

Python实现：

class Solution:    # @param haystack, a string    # @param needle, a string    # @return a string or None    def strStr(self, haystack, needle):        if not needle:            return haystack                len1 = len(haystack)        len2 = len(needle)        if len1 < len2:            return None                for i in range(len1 - len2 + 1):            j = 0            k = i            while j < len2 and needle[j] == haystack[k]:                j += 1                k += 1                        if j == len2:                return haystack[i:]                return None

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