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 findSubstring(String s, String[] words) {
        Map tmp,map=new HashMap();
        int lens=s.length(),lenw=words[0].length();
        for(int i=0;i res = new ArrayList();
        for(int i=0;i<=lens-lenw*words.length;i++){
            tmp=new HashMap();
            int j=0;
            for(;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=length)res[(*returnSize)++]=i;
	}
	for(i=0;i

C++源代码（704ms）：

class Solution {
public:
    vector findSubstring(string s, vector& words) {
        int lens=s.size(),lenw=words[0].size(),length=words.size();
        map strmap,tmp;
        for(int i=0;i res;
        for(int i=0;i=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