[187] C language brush LeetCode repeated DNA sequence (M)

【

All DNA by a series of abbreviated as A, C, G, and T nucleotides, for example: "ACGAATTCCG". In the study of DNA, DNA repeat sequence recognition in the research will be very helpful sometimes.

A write function to find more than once a sequence of 10 characters long (substring) the DNA molecule in all occurrences.

 

Example:

Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
Output: [ "AAAAACCCCC", "CCCCCAAAAA "]

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/repeated-dna-sequences
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

】

This question by treatment with 31/32 cases, and finally with a timeout embodiment. Bit by bit hash estimate should do some first recorded version of this use case does not pass than it will save a hash string.

This question is just to see would feel unable to start. Then look to see the number of such fields, should think of is to use a hash.

So how correspondence between the hash key structure it? Thinking is traversed in the string s, if not recorded, if there will count by one. I.e., the keys are equal, a further need to increase the count array.

This problem to use string comparison function strcmp, string copy function strncpy and strcpy.

Then a large part of the application code is in the malloc memory, C language brush title, this point still have to endure.

 

 

int count;

void Checkhash(char *temp, char **hashstr, char *countarr) {
    int i;

    for (i = 0; i < count; i++) {
        if (!strcmp(temp, hashstr[i])) {
            countarr[i] = 1; // 找到重复
            return;
        }
    }

    strcpy(hashstr[count], temp);
    count++;
}

char ** findRepeatedDnaSequences(char * s, int* returnSize){
    int i;
    char **retarr;
    int len = strlen(s);
    char **hashstr;
    char temp[11] = {0};
    char countarr[len + 1];
    int retcount = 0;

    count = 0;
    
    memset(countarr, 0, (len + 1));

    hashstr = (char **)malloc(sizeof(char *) * len);
    for (i = 0; i < len; i++) {
        hashstr[i] = (char *)malloc(sizeof(char) * 11);
        memset(hashstr[i], 0 , sizeof(char) * 11);
    }

    retarr = (char **)malloc(sizeof(char *) * len);
    for (i = 0; i < len; i++) {
        retarr[i] = (char *)malloc(sizeof(char) * 11);
    }

    for (i = 0; i < len - 9; i++) {
        strncpy(temp, s + i, 10);
        Checkhash(temp, hashstr, countarr);
    }

    for (i = 0 ; i < count; i++) {
        if (countarr[i] > 0) {
            strcpy(retarr[retcount], hashstr[i]);
            retcount++;
        }
    }

    *returnSize = retcount;

    return retarr;
}