練KMP我認爲沒什麼技巧,背代碼就完事了。我的口訣:(僅限於自己理解)
(1)、先回溯主串後回溯子串,回溯哪個哪個爲負一;
(2)、推nextTable時拿子串對應的下標開刀。
PS:這題把nextTable數組換成next數組居然有歧義,這辣雞編譯器居然還會把next數組設成關鍵字,第一次碰到這種錯害得我檢查了一個小時,真是絕了。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <climits>
using namespace std;
const int MAXN = 1000005;
const int INF = INT_MAX;
int nextTable[MAXN];
void getNext(string &text, string &pattern){
int n = text.size();
int m = pattern.size();
int i, j;
j = 0;
nextTable[j] = -1;
i = nextTable[j];
while(j < m){
if(i == -1 || pattern[i] == pattern[j]){
i++;
j++;
nextTable[j] = i;
}
else{
i = nextTable[i];
}
}
}
int KMP(string &text, string &pattern){
getNext(text, pattern);
int n = text.size();
int m = pattern.size();
int i, j, ans;
i = j = ans = 0;
while(i < n && j < m){
if(j == -1 || text[i] == pattern[j]){
i++;
j++;
}
else{
j = nextTable[j];
}
if(j == m){
ans++;
j = nextTable[j];
}
}
return ans;
}
int main(){
// freopen("in.txt", "r", stdin);
string text, pattern;
while(cin >> text >> pattern){
printf("%d\n", KMP(text, pattern));
}
return 0;
}
連續兩個遇到牛客的辣雞題,上一個是編譯器的問題,這個是題目沒說清楚。明明題目中說了"並且可以有一個用中括號表示的模式匹配",是隻有一個括號,結果後來卡在688那個例子傻了半天,無奈看評論發現不只有一個括號。nnd,只能重寫一遍匹配了。有點不舒服,A了是運氣。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <cctype>
#include <climits>
using namespace std;
const int MAXN = 1005;
const int INF = INT_MAX;
int main(){
// freopen("in.txt", "r", stdin);
int n;
string pattern;
string table[MAXN], tabletmp[MAXN];
while(~scanf("%d", &n)){
//if(n == 0) break;
for(int i = 0; i < n; i++){
cin >> table[i];
}
cin >> pattern;
//初始化
for(int i = 0; i < n; i++){
tabletmp[i] = table[i];
for(int j = 0; j < tabletmp[i].size(); j++){
if(isupper(tabletmp[i][j])) tabletmp[i][j] = tolower(tabletmp[i][j]);
}
}
for(int i = 0; i < pattern.size(); i++){
if(isupper(pattern[i])) pattern[i] = tolower(pattern[i]);
}
//模式串匹配
for(int i = 0; i < n; i++){
bool flag1 = true;
int num = 0;
for(int j = 0; j < tabletmp[i].size(); j++){
if(pattern[num] == '['){
if(pattern[num+1] == ']') continue;
bool flag2 = false;//初始爲匹配不到
num++;
while(pattern[num] != ']'){
if(pattern[num] == tabletmp[i][j]) flag2 = true;//匹配到
num++;
}
if(!flag2){
flag1 = false;
break;
}
num++;
continue;
}
if(pattern[num] != tabletmp[i][j]){
flag1 = false;
break;
}
num++;
}
if(num != pattern.size()) flag1 = false;
if(flag1) cout << i+1 << " " << table[i] << endl;
}
}
return 0;
}