问题:"火柴堆找针”[1]
输入:“火柴堆”string haystack,“针”string needle
输出:火柴堆中第一根针出现的位置
思路:
two pointer，外层循环遍历火柴堆，内层循环遍历针
代码如下:
class Solution {
public:
int strStr(string haystack, string needle) {
int i,j,k,h_l,n_l;
int index;
h_l=haystack.length();
n_l=needle.length();
if(n_l==0)
{
return 0;
}
for(i=0;i<(h_l-n_l+1);i++)
{
k=i; //needle起始位置
for(j=0;j<n_l;j++)
{
if(haystack[k++]==needle[j])
{
continue;
}
else
{
break;
}
}
if(j==n_l)
{
return i;
}
}
return -1;
}
};运行结果:
Runtime: 8 ms
Memory Usage: 9.5 MB
围观了隔壁大神的solution，发现了Knuth Morris Pratt的知识点，学习一下~[2]
KMP用于在text[0..n-1] 中，搜索pattern[0..m-1]，最坏时间复杂度为O(n)。
具体思想：KMP使用的是pattern的退化性质（有些sub-pattern在pattern中出现不止一次 ）
无论何时检测到mismatch，我们已经知道下个窗口的text的一些字符，利用这个信息避免匹配那些（我们已知）无论如何都会匹配的字符。
看个例子，进一步了解KMP
Matching Overview txt = "AAAAABAAABA" pat = "AAAA" We compare first window of txt with pat txt = "AAAAABAAABA" pat = "AAAA" [Initial position] We find a match. This is same as Naive String Matching. In the next step, we compare next window of txt with pat. txt = "AAAAABAAABA" pat = "AAAA" [Pattern shifted one position] This is where KMP does optimization over Naive. In this second window, we only compare fourth A of pattern with fourth character of current window of text to decide whether current window matches or not. Since we know first three characters will anyway match, we skipped matching first three characters.Need of Preprocessing? An important question arises from the above explanation, how to know how many characters to be skipped. To know this, we pre-process pattern and prepare an integer array lps[] that tells us the count of characters to be skipped.
Preprocessing Overview:
- KMP algorithm preprocesses pat[] and constructs an auxiliary lps[] of size m (same as size of pattern) which is used to skip characters while matching.
- name lps indicates longest proper prefix which is also suffix.. A proper prefix is prefix with whole string not allowed. For example, prefixes of “ABC” are “”, “A”, “AB” and “ABC”. Proper prefixes are “”, “A” and “AB”. Suffixes of the string are “”, “C”, “BC” and “ABC”.
- We search for lps in sub-patterns. More clearly we focus on sub-strings of patterns that are either prefix and suffix.
- For each sub-pattern pat[0..i] where i = 0 to m-1, lps[i] stores length of the maximum matching proper prefix which is also a suffix of the sub-pattern pat[0..i].
lps[i] = the longest proper prefix of pat[0..i] which is also a suffix of pat[0..i].
Note : lps[i] could also be defined as longest prefix which is also proper suffix. We need to use properly at one place to make sure that the whole substring is not considered.
Examples of lps[] construction: For the pattern “AAAA”, lps[] is [0, 1, 2, 3] For the pattern “ABCDE”, lps[] is [0, 0, 0, 0, 0] For the pattern “AABAACAABAA”, lps[] is [0, 1, 0, 1, 2, 0, 1, 2, 3, 4, 5] For the pattern “AAACAAAAAC”, lps[] is [0, 1, 2, 0, 1, 2, 3, 3, 3, 4] For the pattern “AAABAAA”, lps[] is [0, 1, 2, 0, 1, 2, 3]在pattern[0..i]中，lps存放着既是前缀又是后缀的子字符串(不包含pattern[0..i]本身)的最大长度
Searching Algorithm:
Unlike Naive algorithm, where we slide the pattern by one and compare all characters at each shift, we use a value from lps[] to decide the next characters to be matched. The idea is to not match a character that we know will anyway match.
How to use lps[] to decide next positions (or to know a number of characters to be skipped)?
- We start comparison of pat[j] with j = 0 with characters of current window of text.
- We keep matching characters txt[i] and pat[j] and keep incrementing i and j while pat[j] and txt[i] keep matching.
- When we see a mismatch
- We know that characters pat[0..j-1] match with txt[i-j…i-1] (Note that j starts with 0 and increment it only when there is a match).
- We also know (from above definition) that lps[j-1] is count of characters of pat[0…j-1] that are both proper prefix and suffix.
- From above two points, we can conclude that we do not need to match these lps[j-1] characters with txt[i-j…i-1] because we know that these characters will anyway match. Let us consider above example to understand this.
txt[] = "AAAAABAAABA" pat[] = "AAAA" lps[] = {0, 1, 2, 3} i = 0, j = 0 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] match, do i++, j++ i = 1, j = 1 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] match, do i++, j++ i = 2, j = 2 txt[] = "AAAAABAAABA" pat[] = "AAAA" pat[i] and pat[j] match, do i++, j++ i = 3, j = 3 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] match, do i++, j++ i = 4, j = 4 Since j == M, print pattern found and reset j, j = lps[j-1] = lps[3] = 3 Here unlike Naive algorithm, we do not match first three characters of this window. Value of lps[j-1] (in above step) gave us index of next character to match. i = 4, j = 3 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] match, do i++, j++ i = 5, j = 4 Since j == M, print pattern found and reset j, j = lps[j-1] = lps[3] = 3 Again unlike Naive algorithm, we do not match first three characters of this window. Value of lps[j-1] (in above step) gave us index of next character to match. i = 5, j = 3 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] do NOT match and j > 0, change only j j = lps[j-1] = lps[2] = 2 i = 5, j = 2 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] do NOT match and j > 0, change only j j = lps[j-1] = lps[1] = 1 i = 5, j = 1 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] do NOT match and j > 0, change only j j = lps[j-1] = lps[0] = 0 i = 5, j = 0 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] do NOT match and j is 0, we do i++. i = 6, j = 0 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] match, do i++ and j++ i = 7, j = 1 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] match, do i++ and j++ We continue this way...
// C++ program for implementation of KMP pattern searching
// algorithm
#include <bits/stdc++.h>
void computeLPSArray(char* pat, int M, int* lps);
// Prints occurrences of txt[] in pat[]
void KMPSearch(char* pat, char* txt)
{
int M = strlen(pat);
int N = strlen(txt);
// create lps[] that will hold the longest prefix suffix
// values for pattern
int lps[M];
// Preprocess the pattern (calculate lps[] array)
computeLPSArray(pat, M, lps);
int i = 0; // index for txt[]
int j = 0; // index for pat[]
while (i < N) {
if (pat[j] == txt[i]) {
j++;
i++;
}
if (j == M) {
printf("Found pattern at index %d ", i - j);
j = lps[j - 1];
}
// mismatch after j matches
else if (i < N && pat[j] != txt[i]) {
// Do not match lps[0..lps[j-1]] characters,
// they will match anyway
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
}
// Fills lps[] for given patttern pat[0..M-1]
void computeLPSArray(char* pat, int M, int* lps)
{
// length of the previous longest prefix suffix
int len = 0;
lps[0] = 0; // lps[0] is always 0
// the loop calculates lps[i] for i = 1 to M-1
int i = 1;
while (i < M) {
if (pat[i] == pat[len]) {
len++;
lps[i] = len;
i++;
}
else // (pat[i] != pat[len])
{
// This is tricky. Consider the example.
// AAACAAAA and i = 7. The idea is similar
// to search step.
if (len != 0) {
len = lps[len - 1];
// Also, note that we do not increment
// i here
}
else // if (len == 0)
{
lps[i] = 0;
i++;
}
}
}
}
// Driver program to test above function
int main()
{
char txt[] = "ABABDABACDABABCABAB";
char pat[] = "ABABCABAB";
KMPSearch(pat, txt);
return 0;
} output：
Found pattern at index 10
Preprocessing Algorithm:
In the preprocessing part, we calculate values in lps[]. To do that, we keep track of the length of the longest prefix suffix value (we use len variable for this purpose) for the previous index. We initialize lps[0] and len as 0. If pat[len] and pat[i] match, we increment len by 1 and assign the incremented value to lps[i]. If pat[i] and pat[len] do not match and len is not 0, we update len to lps[len-1]. See computeLPSArray () in the below code for details.
Illustration of preprocessing (or construction of lps[])
pat[] = "AAACAAAA" len = 0, i = 0. lps[0] is always 0, we move to i = 1 len = 0, i = 1. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 1, lps[1] = 1, i = 2 len = 1, i = 2. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 2, lps[2] = 2, i = 3 len = 2, i = 3. Since pat[len] and pat[i] do not match, and len > 0, set len = lps[len-1] = lps[1] = 1 len = 1, i = 3. Since pat[len] and pat[i] do not match and len > 0, len = lps[len-1] = lps[0] = 0 len = 0, i = 3. Since pat[len] and pat[i] do not match and len = 0, Set lps[3] = 0 and i = 4. We know that characters pat len = 0, i = 4. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 1, lps[4] = 1, i = 5 len = 1, i = 5. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 2, lps[5] = 2, i = 6 len = 2, i = 6. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 3, lps[6] = 3, i = 7 len = 3, i = 7. Since pat[len] and pat[i] do not match and len > 0, set len = lps[len-1] = lps[2] = 2 len = 2, i = 7. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 3, lps[7] = 3, i = 8 We stop here as we have constructed the whole lps[].
使用KMP解决本题，代码如下：
class Solution {
public:
vector<int> lps(string needle)
{
int n=needle.size();
vector<int> l(n,0);
for(int i=1,len=0;i<n;)
{
if(needle[i]==needle[len])
{
l[i++]=++len;
}
else if(len)
{
len=l[len-1];
}
else
{
l[i++]=0;
}
}
return l;
}
int strStr(string haystack, string needle) {
int m=haystack.size();
int n=needle.size();
if(!n)
{
return 0;
}
vector<int> l=lps(needle);
for(int i=0,j=0;i<m;)
{
if(haystack[i]==needle[j])
{
i++,j++;
}
if(j==n)
{
return i-j;
}
if(i<m&&haystack[i]!=needle[j])
{
j?j=l[j-1]:i++;
}
}
return -1;
}
};Runtime: 8 ms
Memory Usage: 9.7 MB
和暴力差不多。
