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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> >= 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;
else
return 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