leetcode28. 实现strStr() 🌟

题目：

　　实现 strStr() 函数。

　　给定一个 haystack 字符串和一个 needle 字符串，在 haystack 字符串中找出 needle 字符串出现的第一个位置 (从0开始)。如果不存在，则返回  -1。

示例 1:

　　输入: haystack = "hello", needle = "ll"
　　输出: 2
示例 2:

　　输入: haystack = "aaaaa", needle = "bba"
　　输出: -1
说明:

　　当 needle 是空字符串时，我们应当返回什么值呢？这是一个在面试中很好的问题。

　　对于本题而言，当 needle 是空字符串时我们应当返回 0 。这与C语言的 strstr() 以及 Java的 indexOf() 定义相符。

来源：力扣（LeetCode）
解答：

leetcode优秀方案(来自力扣答案统计页，没有明确作者是谁，可留言告知)：

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        for i in range(len(haystack) - len(needle) + 1):
            if haystack[i: i + len(needle)] == needle:
                return i
        return -1

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        if needle == '':
            return 0

        i = 0
        while i <= len(haystack) - len(needle):
            if haystack[i] != needle[0]:
                i += 1
            else:
                j = i + 1
                k = 1
                while k < len(needle):
                    if haystack[j] == needle[k]:
                        j += 1
                        k += 1
                    else:
                        i += 1
                        break
                else:
                    return i
        else:
            return -1

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        if len(needle) == 0:
            return 0
        i = 0
        j = 0
        while i < len(haystack) and j < len(needle):
            if haystack[i] == needle[j]:
                i += 1
                j += 1
            else:
                i = i - j + 1
                j = 0
        if j == len(needle):
            return i - j
        return -1
    # https://leetcode-cn.com/problems/implement-strstr/comments/53265

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        if needle == '': return 0
        for i in range(len(haystack) - len(needle) + 1):
            if haystack[i] == needle[0]:
                first = ''
                for j in range(len(needle)):
                    first += haystack[i+j]
                if first == needle:
                    return i
        
        return -1

# 作者：jian-chuan
# 链接：https://leetcode-cn.com/problems/two-sum/solution/nei-zhi-han-shu-jiu-bu-xie-liao-lai-ge-bao-li-po-j/

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        if not needle : return 0
        _next = [-1]
        
        def getNext(p, _next):
            _next[0] = -1
            i = 0
            j = -1
            while i < len(p) - 1:
                if j == -1 or p[i] == p[j]:
                    i += 1
                    j += 1
                    # _next.append(j)
                    if p[i] != p[j]:
                        _next.append(j)
                    else:
                        _next.append(_next[j])
                else:
                    j = _next[j]
        getNext(needle, _next)
        
        i = 0
        j = 0
        while i < len(haystack) and j < len(needle):
            if j == -1 or haystack[i] == needle[j]:
                i += 1
                j += 1
            else:
                j = _next[j]
        if j == len(needle):
            return i - j
        return -1

# 作者：powcai
# 链接：https://leetcode-cn.com/problems/two-sum/solution/shi-xian-strstr-by-powcai/
# 链接：https://leetcode-cn.com/problems/implement-strstr/solution/c5chong-jie-fa-ku-han-shu-bfkmpbmsunday-by-2227

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        def get_bmB(T, bmB):
            tlen = len(T)
            for i in range(256):
                bmB.append(len(T))
            for i in range(tlen - 1):
                bmB[ord(T[i])] = tlen - i - 1
        
        def get_suff(T, suff):
            tlen = len(T)
            for i in range(tlen - 2 , -1 , -1):
                k = i
                while k > 0 and T[k] == T[tlen - 1 - i + k]:
                    k -= 1
                suff[i] = i - k
        
        def get_bmG(T, bmG):
            tlen = len(T)
            suff = [0] * (tlen + 1)
            get_suff(T, suff)
            for i in range(tlen):
                bmG[i] = tlen
            for i in range(tlen - 1, -1, -1):
                if suff[i] == i + 1:
                    for j in range(tlen - 1):
                        if bmG[j] == tlen:
                            bmG[j] = tlen - 1 - i
            for i in range(tlen - 1):
                bmG[tlen - 1 - suff[i]] = tlen - 1 - i
        
        i = 0
        tlen = len(needle)
        slen = len(haystack)
        bmG = [0] * tlen
        bmB = []
        get_bmB(needle, bmB)
        get_bmG(needle, bmG)
        
        while i <= slen - tlen:
            j = tlen - 1
            while j > -1 and haystack[i + j] == needle[j]:
                    j -= 1
            if j == -1:
                return i
            
            i += max(bmG[j], bmB[ord(haystack[i+j])] - (tlen - 1 - j))
        
        return -1

# https://leetcode-cn.com/problems/implement-strstr/solution/c5chong-jie-fa-ku-han-shu-bfkmpbmsunday-by-2227

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        if needle == '': return 0
        
        slen = len(haystack)
        tlen = len(needle)
        if slen < tlen: return -1
        
        i = j = 0
        m = tlen
        
        while i < slen:
            if haystack[i] != needle[j]:
                for k in range(tlen - 1, -1, -1):
                    if m < slen and needle[k] == haystack[m]:
                        break
                i = m - k
                j = 0
                m = i + tlen
                if m > slen:
                    return -1
            else:
                if j == tlen - 1:
                    return i - j
                i += 1
                j += 1
        
        return -1

# https://leetcode-cn.com/problems/implement-strstr/solution/c5chong-jie-fa-ku-han-shu-bfkmpbmsunday-by-2227

字符串匹配的KMP算法
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