# strStr

## Question

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.


## 題解

### 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;
}
}


### 源碼分析

1. 邊界檢查：sourcetarget有可能是空串。
2. 邊界檢查之下標溢出：注意變量i的循環判斷條件，如果是單純的i < source.length()則在後面的source.charAt(i + j)時有可能溢出。
3. 代碼風格：（1）運算符==兩邊應加空格；（2）變量名不要起s1s2這類，要有意義，如targetsource；（3）即使if語句中只有一句話也要加大括號，即{return -1;}；（4）Java 代碼的大括號一般在同一行右邊，C++ 代碼的大括號一般另起一行；（5）int i, j;聲明前有一行空格，是好的代碼風格。
4. 不要在for的條件中聲明i,j，容易在循環外再使用時造成編譯錯誤，錯誤代碼示例：

## 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;
}
}