题目描述:
Implement strStr().
Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
Example 1:
Input: haystack = "hello", needle = "ll"
Output: 2
Example 2:
Input: haystack = "aaaaa", needle = "bba"
Output: -1
Clarification:
What should we return when needle is an empty string? This is a great question to ask during an interview.
For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().
解题思路:
1,滑动窗口的思想:
把needle当成一个整体,滑动窗口size为needle.length(),在haystack中按照每次一个字符的速度去滑动;
比如:当haystack="helloworld",needle="low"时,第一次滑动,比较"hel"与"low",第二次比较"ell"与"low",第三次比较"llo"与"low",如此类推下去;
这个过程只需要一个for循环即可,for循环终止条件为haystack.length()-needle.length()+1;for循环有两种输出可能,第一,成功匹配,则返回对应索引。第二,为匹配,则返回-1;
接下来的重点是判断size为needle.length()的滑动窗口的字符串与needle是否相等,这个过程也同样只需要一个for循环即可,for循环结束条件为needle.length(),这一层的主要思想就是利用了for循环的break,当第一个字符不相等时,就不需要继续匹配后面的字符,直接break;
2,匹配成功的条件:
index=i+needle.length(),index定义见程序中;
3,结束条件:
滑动至haystack.length()-needle.length()+1时结束;
难点:
笔者在做这道题的时候提交了许多次,错误大概有needle为空与haystack为空考虑不周全,所以还是需要细心
代码:
class Solution {
public int strStr(String haystack, String needle) {
if(haystack.length() < needle.length()){
return -1;
}
else if(haystack.length() ==0 && needle.length()==0){
return 0;
}
else {
for (int i = 0; i < haystack.length() - needle.length() + 1; i++) {
int index = i;
for (int j = 0; j < needle.length(); j++) {
if (haystack.charAt(index)!=needle.charAt(j)){
break;
}
index += 1;
}
if(index == i + needle.length()){
return i;
}
}
return -1;
}
}
}
其他方法:
当然这道题目也有其它优秀的方法,比如下面这个只用了一行代码,但这个显然不是这道题目的本意,忘读者脚踏实地;
class Solution {
public int strStr(String haystack, String needle) {
if(haystack.contains(needle)){
return haystack.indexOf(needle);
}
return -1;
}
}