LeetCode-28 实现strStr()

文章目录

题目描述
我的解法
反思
优化
其他思路
总结
Github

题目描述

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。

我的解法

public int strStr(String haystack, String needle) {//特殊情况if(null == needle || "".equals(needle))return 0;if(null == haystack || "".equals(haystack)||!haystack.contains(needle))return -1;//接下来肯定是haystack包含needle的情况char needleFirstChar = needle.charAt(0);//获取needle的第一位字符int returnInt = haystack.indexOf(needleFirstChar);//找到needle第一位字符在haystack中第一次出现的位置，初始化String partHaystack = haystack.substring(returnInt,returnInt+needle.length());//partHaystack是从returnInt开始截取needle长度的字符串while(!needle.equals(partHaystack)){//先判断partHaystack是否跟指定字符串相等returnInt++;returnInt += haystack.substring(returnInt).indexOf(needleFirstChar);//取从returnInt之后以为到末尾组成的新字符串中，第一次出现needleFirstChar的位置partHaystack = haystack.substring(returnInt,returnInt+needle.length());//从该位置截取needle长度的字符串用于作比较}return returnInt;}

用间：5ms
战胜：90.31%

反思

其实String类的contains方法就提供了一个参数为String的类似的方法。不过参数不能为空，否则会报空指针异常“NullPointerException”

 /*** Returns the index within this string of the first occurrence of the* specified substring.** <p>The returned index is the smallest value <i>k</i> for which:* <blockquote><pre>* this.startsWith(str, <i>k</i>)* </pre></blockquote>* If no such value of <i>k</i> exists, then {@code -1} is returned.** @param   str   the substring to search for.* @return  the index of the first occurrence of the specified substring,*          or {@code -1} if there is no such occurrence.*/public int indexOf(String str) {return indexOf(str, 0);}/*** Returns the index within this string of the first occurrence of the* specified substring, starting at the specified index.** <p>The returned index is the smallest value <i>k</i> for which:* <blockquote><pre>* <i>k</i> &gt;= fromIndex {@code &&} this.startsWith(str, <i>k</i>)* </pre></blockquote>* If no such value of <i>k</i> exists, then {@code -1} is returned.** @param   str         the substring to search for.* @param   fromIndex   the index from which to start the search.* @return  the index of the first occurrence of the specified substring,*          starting at the specified index,*          or {@code -1} if there is no such occurrence.*/public int indexOf(String str, int fromIndex) {return indexOf(value, 0, value.length,str.value, 0, str.value.length, fromIndex);}
/*** Code shared by String and StringBuffer to do searches. The* source is the character array being searched, and the target* is the string being searched for.** @param   source       the characters being searched.* @param   sourceOffset offset of the source string.* @param   sourceCount  count of the source string.* @param   target       the characters being searched for.* @param   targetOffset offset of the target string.* @param   targetCount  count of the target string.* @param   fromIndex    the index to begin searching from.*/static int indexOf(char[] source, int sourceOffset, int sourceCount,char[] target, int targetOffset, int targetCount,int fromIndex) {if (fromIndex >= sourceCount) {return (targetCount == 0 ? sourceCount : -1);}if (fromIndex < 0) {fromIndex = 0;}if (targetCount == 0) {return fromIndex;}char first = target[targetOffset];int max = sourceOffset + (sourceCount - targetCount);for (int i = sourceOffset + fromIndex; i <= max; i++) {/* Look for first character. */if (source[i] != first) {while (++i <= max && source[i] != first);}/* Found first character, now look at the rest of v2 */if (i <= max) {int j = i + 1;int end = j + targetCount - 1;for (int k = targetOffset + 1; j < end && source[j]== target[k]; j++, k++);if (j == end) {/* Found whole string. */return i - sourceOffset;}}}return -1;}

优化

public int strStr1(String haystack, String needle) {//特殊情况if(null == needle || "".equals(needle))return 0;elsereturn haystack.indexOf(needle);}

用时：6ms（可能电脑及网络性能影响）
战胜：80.45%

其他思路

我最先想到可以先找到needle首个字符在haystack的位置，在用for循环遍历haystack这个字符串needle的长度位的字符，让他们一一比较，如果不匹配则再找到下一个与needle首个字符一致的字符位置，接着遍历。这样可以省去重复建造字符串所导致的系统开销，从而稍稍能缩短运行时间。但其实String类原生的实现方法跟这种方法差不多，具体可以参考一下人家写的代码。

总结

while循环是先判断后循环，而do–while循环是先循环后判断
关于substring的双参方法substring(a, b),指的是截取字符串 [a,b)左闭右开位字符

The substring begins at the specified {@code beginIndex} and
extends to the character at index {@code endIndex - 1}.
Thus the length of the substring is {@code endIndex-beginIndex}
还需多看jdk源码

Github

LeetCode刷题笔记