strStr
最后更新于：2022-04-02 01:07:33
# strStr
### Source
- leetcode: [Implement strStr() | LeetCode OJ](https://leetcode.com/problems/implement-strstr/)
- lintcode: [lintcode - (13) strstr](http://www.lintcode.com/zh-cn/problem/strstr/)
~~~
strstr (a.k.a find sub string), is a useful function in string operation.
You task is to implement this function.
For a given source string and a target string,
you should output the "first" index(from 0) of target string in source string.
If target is not exist in source, just return -1.
Example
If source="source" and target="target", return -1.
If source="abcdabcdefg" and target="bcd", return 1.
Challenge
O(n) time.
Clarification
Do I need to implement KMP Algorithm in an interview?
- Not necessary. When this problem occurs in an interview,
the interviewer just want to test your basic implementation ability.
~~~
### 题解
对于字符串查找问题，可使用双重for循环解决，效率更高的则为KMP算法。
### Java
~~~
/**
* http://www.jiuzhang.com//solutions/implement-strstr
*/
class Solution {
/**
* Returns a index to the first occurrence of target in source,
* or -1 if target is not part of source.
* @param source string to be scanned.
* @param target string containing the sequence of characters to match.
*/
public int strStr(String source, String target) {
if (source == null || target == null) {
return -1;
}
int i, j;
for (i = 0; i < source.length() - target.length() + 1; i++) {
for (j = 0; j < target.length(); j++) {
if (source.charAt(i + j) != target.charAt(j)) {
break;
} //if
} //for j
if (j == target.length()) {
return i;
}
} //for i
// did not find the target
return -1;
}
}
~~~
### 源码分析
1. 边界检查：`source`和`target`有可能是空串。
1. 边界检查之下标溢出：注意变量`i`的循环判断条件，如果是单纯的`i < source.length()`则在后面的`source.charAt(i + j)`时有可能溢出。
1. 代码风格：（1）运算符`==`两边应加空格；（2）变量名不要起`s1``s2`这类，要有意义，如`target``source`；（3）即使if语句中只有一句话也要加大括号，即`{return -1;}`；（4）Java 代码的大括号一般在同一行右边，C++ 代码的大括号一般另起一行；（5）`int i, j;`声明前有一行空格，是好的代码风格。
1. 不要在for的条件中声明`i`,`j`，容易在循环外再使用时造成编译错误，错误代码示例：
### Another Similar Question
~~~
/**
* http://www.jiuzhang.com//solutions/implement-strstr
*/
public class Solution {
public String strStr(String haystack, String needle) {
if(haystack == null || needle == null) {
return null;
}
int i, j;
for(i = 0; i < haystack.length() - needle.length() + 1; i++) {
for(j = 0; j < needle.length(); j++) {
if(haystack.charAt(i + j) != needle.charAt(j)) {
break;
}
}
if(j == needle.length()) {
return haystack.substring(i);
}
}
return null;
}
}
~~~
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