strStr
Question
- leetcode: Implement strStr() | LeetCode OJ
- lintcode: lintcode - (13) 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.
題解
對於字串查找問題,可使用雙重for迴圈解決,效率更高的則為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;
}
}
源碼分析
- 邊界檢查:
source
和target
有可能是空串。 - 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。 - 代碼風格:(1)運算符
==
兩邊應加空格;(2)變量名不要起s1``s2
這類,要有意義,如target``source
;(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。 - 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;
}
}