leetCode(30) - Substring with Concatenation of All Words

最新推荐文章于 2020-05-30 14:27:40 发布

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本文介绍了一种用于查找字符串中特定模式的算法。该算法能够高效地找出由一组固定长度单词构成的所有组合出现的位置。通过使用哈希表进行单词匹配计数，文章提供了C语言和C++实现的详细代码示例。

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Problem:

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in wordsexactly once and without any intervening characters.

For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

Solution:

由于长度是固定的，因此固定每一个字符串开始的位置，然后向后搜索，查看每一个字符串是否存在，如果不存在i++

C语言源代码（260ms）：

sub不free的话会MLE

[cpp] view plain copy

typedef struct Node{
char* word;
int times;
struct Node* next;
}data;
#define SIZE 1000
int hash(char* word){
int i=0,h=0;
for(;word[i];i++){
h=(h*31+word[i])%SIZE;
}
return h;
}
int InsertMap(data** map,char* word,int lenw){
int h = hash(word);
if(map[h]==NULL){
map[h]=(data*)malloc(sizeof(data));
map[h]->word=(char*)malloc(sizeof(char)*(lenw+1));
map[h]->times=1;
strcpy(map[h]->word,word);
map[h]->next=NULL;
return 1;
}else{
data* p=map[h];
while(p->next!=NULL){
if(strcmp(p->word,word)==0){
p->times++;
return p->times;
}
p=p->next;
}
if(strcmp(p->word,word)==0){
p->times++;
return p->times;
}else{
data* tmp=(data*)malloc(sizeof(data));
tmp->word=(char*)malloc(sizeof(char)*(lenw+1));
tmp->times=1;
strcpy(tmp->word,word);
tmp->next=NULL;
p->next=tmp;
return 1;
}
}
}
int FindMap(data** map,char* sub){
int h = hash(sub);
if(map[h]==NULL){
return -1;
}else{
data* p=map[h];
while(p!=NULL){
if(strcmp(p->word,sub)==0){
return p->times;
}
p=p->next;
}
return -1;
}
}
char* substring(char* s,int start,int len){
char* sub=(char*)malloc(sizeof(char)*(len+1));
int i=0;
for(;i<len;i++){
sub[i]=s[i+start];
}
sub[i]=0;
return sub;
}
int* findSubstring(char* s, char** words, int wordsSize, int* returnSize) {
int lenw=strlen(words[0]),lens=strlen(s),length=wordsSize;
int* res=(int*)malloc(sizeof(int)*(lens-lenw*length+1));
data** map=(data**)malloc(sizeof(data*)*SIZE);
data** tmp=(data**)malloc(sizeof(data*)*SIZE);
int i,j;
for(i=0;i<SIZE;i++){
map[i]=NULL;
tmp[i]=NULL;
}
for(i=0;i<length;i++){
InsertMap(map,words[i],lenw);
}
*returnSize=0;
for(i=0;i<lens-lenw*length+1;i++){
for(j=0;j<SIZE;j++){
if(tmp[j]!=NULL){
free(tmp[j]);
tmp[j]=NULL;
}
}
for(j=0;j<length;j++){
char* sub=substring(s,i+j*lenw,lenw);
int mapnum=FindMap(map,sub);
if(mapnum==-1)break;
int num=InsertMap(tmp,sub,lenw);
if(mapnum < num)break;
free(sub);
}
if(j>=length)res[(*returnSize)++]=i;
}
for(i=0;i<SIZE;i++)if(map[i]!=NULL)free(map[i]);
free(map);
return res;
}

C++源代码（704ms）：

[cpp] view plain copy

class Solution {
public:
vector<int> findSubstring(string s, vector<string>& words) {
int lens=s.size(),lenw=words[0].size(),length=words.size();
map<string,int> strmap,tmp;
for(int i=0;i<length;i++){
strmap[words[i]]++;
}
vector<int> res;
for(int i=0;i<lens-lenw*length+1;i++){
tmp.clear();
int j=0;
for(j=0;j<length;j++){
string sub=s.substr(i+j*lenw,lenw);
if(strmap.find(sub) == strmap.end())break;
tmp[sub]++;
if(strmap[sub] < tmp[sub])break;
}
if(j>=length)res.push_back(i);
}
return res;
}
};