LeetCode 30 Substring with Concatenation of All Words (C,C++,Java,Python)

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 1000int 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