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Update 0028.实现strStr.md

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`564`	`564`	`## 其他语言版本`
`565`	`565`
`566`	`566`	`### Java:`
	`567`	+```Java
	`568`	`+class Solution {`
	`569`	`+ /**`
	`570`	`+ 牺牲空间,换取最直白的暴力法`
	`571`	`+ 时间复杂度 O(n * m)`
	`572`	`+ 空间 O(n + m)`
	`573`	`+ */`
	`574`	`+ public int strStr(String haystack, String needle) {`
	`575`	`+ // 获取 haystack 和 needle 的长度`
	`576`	`+ int n = haystack.length(), m = needle.length();`
	`577`	`+ // 将字符串转换为字符数组,方便索引操作`
	`578`	`+ char[] s = haystack.toCharArray(), p = needle.toCharArray();`
	`579`	`+`
	`580`	`+ // 遍历 haystack 字符串`
	`581`	`+ for (int i = 0; i < n - m + 1; i++) {`
	`582`	`+ // 初始化匹配的指针`
	`583`	`+ int a = i, b = 0;`
	`584`	`+ // 循环检查 needle 是否在当前位置开始匹配`
	`585`	`+ while (b < m && s[a] == p[b]) {`
	`586`	`+ // 如果当前字符匹配,则移动指针`
	`587`	`+ a++;`
	`588`	`+ b++;`
	`589`	`+ }`
	`590`	`+ // 如果 b 等于 m,说明 needle 已经完全匹配,返回当前位置 i`
	`591`	`+ if (b == m) return i;`
	`592`	`+ }`
	`593`	`+`
	`594`	`+ // 如果遍历完毕仍未找到匹配的子串,则返回 -1`
	`595`	`+ return -1;`
	`596`	`+ }`
	`597`	`+}`
	`598`	+```
`567`	`599`
`568`	`600`	```Java
`569`	`601`	`class Solution {`

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