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++
Java源代码（658ms）：
public class Solution {
public List<Integer> findSubstring(String s, String[] words) {
Map<String,Integer> tmp,map=new HashMap<String,Integer>();
int lens=s.length(),lenw=words[0].length();
for(int i=0;i<words.length;i++){
if(map.containsKey(words[i])){
map.put(words[i],map.get(words[i])+1);
}else{
map.put(words[i],1);
}
}
List<Integer> res = new ArrayList<Integer>();
for(int i=0;i<=lens-lenw*words.length;i++){
tmp=new HashMap<String,Integer>();
int j=0;
for(;j<words.length;j++){
int pos=i+j*lenw;
String sub=s.substring(pos,pos+lenw);
if(map.containsKey(sub)){
int num=0;
if(tmp.containsKey(sub))num=tmp.get(sub);
if(map.get(sub) < num+1)break;
else tmp.put(sub,num+1);
}else break;
}
if(j>=words.length){
res.add(i);
}
}
return res;
}
}
C语言源代码（260ms）：
sub不free的话会MLE
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）：
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;
}
};
Python源代码（706ms）：
class Solution:
# @param {string} s
# @param {string[]} words
# @return {integer[]}
def findSubstring(self, s, words):
lens=len(s);lenw=len(words[0]);length=len(words)
map={};res=[]
for i in range(length):
if words[i] in map:map[words[i]]+=1
else:map[words[i]]=1
if not words[i] in s:return res
for i in range(lens-lenw*length+1):
tmp={};j=0;flag=True
for j in range(length):
pos=i+j*lenw
sub=s[pos:pos+lenw]
if sub in map:
num=0
if sub in tmp:num=tmp[sub]
if map[sub] < num+1:flag=False;break
tmp[sub]=num+1
else:flag=False;break
if flag:res.append(i)
return res