strstr (a.k.a find sub string), is a useful function in string operation. Your task is to implement this function.
For a given source string and a target string, you should output the firstindex(from 0) of target string in source string.
If target does not exist in source, just return -1
.
If source = "source"
and
target = "target"
, return -1
.
If source = "abcdabcdefg"
and
target = "bcd"
, return 1
.
O(n2) is acceptable. Can you implement an O(n) algorithm? (hint: KMP)
Do I need to implement KMP Algorithm in a real interview?
- Not necessary. When you meet this problem in a real interview, the interviewer may just want to test your basic implementation ability. But make sure your confirm with the interviewer first.
題目至少有3種方法,第一種是暴力搜索,每次移動一步,則時間複雜度爲O(nm),其中n表示source的長度,m表示target的長度。第二種方法是在第一種方法的基礎上做了改進,每次對比時,同時從兩邊向中間夾,這樣可以適當降低時間複雜度;第三種方法,肯定就是KMP了,KMP算法優化的是第一二種方法存在的回溯問題,第一二種方法中,每次移動i到達某個值,使得source[i] != target[j]的時候,會回溯到i - j + 1的位置,也就是說[i-j+1, i]這個區間又要重複計算;
KMP算法的提出就是要避免此回溯,使得i只往後移,也就是說,這樣的時間複雜度將是O(N),N爲source的長度。KMP的算法證明,就不詳細展開了;這裏主要講下next[j]的求解,其實在求解next[j]的過程有遞歸的含義在其中,也是在做一個字符串的匹配問題,有兩個變量k,j,分別初始化爲k = -1, j = 0,next[0] = -1;當k == -1 或者 target[k] == target[j]的時候,則說明next[j + 1] = k + 1; 否則的話,則將k移動到next[k],這一步就包括了遞推的字符串匹配在其中,理解起來比較難;
class Solution {
public:
/**
* 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.
*/
//KMP算法
void GetNext(const char *p, vector<int> &next)
{
int len = strlen(p);
int k = -1, j = 0;
while(j < len - 1)
{
if(k == -1 || p[k] == p[j])
{
next[++j] = ++k;
}
else
k = next[k];
}
}
int strStr(const char *source, const char *target) {
// write your code here
if(source == NULL || target == NULL)
return -1;
int lens = strlen(source);
int lent = strlen(target);
if(lens < lent)
return -1;
int i = 0, j = 0;
vector<int> next(lent, -1);
GetNext(target, next);
while(i < lens && j < lent)
{
if(j == -1)
{
++i;
j = 0;
}
else if(source[i] == target[j])
{
++i;
++j;
}
else
{
j = next[j];
}
}
if(j >= lent)
return i - j;
return -1;
}
};