LeetCode-Repeated DNA Sequences (位图算法减少内存)

Repeated DNA Sequences

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA.

Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.

For example,

Given s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT",

Return:
["AAAAACCCCC", "CCCCCAAAAA"].

 

用位图算法可以减少内存，代码如下：

int map_exist[ *  / ];
int map_pattern[ *  / ];

#define set(map,x) \
    (map[x >> ] |= ( << (x & 0x1F)))

#define test(map,x) \
    (map[x >> ] & ( << (x & 0x1F)))

int dnamap[];

char** findRepeatedDnaSequences(char* s, int* returnSize) {
    *returnSize = ;
    if (s == NULL) return NULL;
    int len = strlen(s);
    if (len <= ) return NULL;

    memset(map_exist, , sizeof(int)* ( *  / ));
    memset(map_pattern, , sizeof(int)* ( *  / ));

    dnamap['A' - 'A'] = ;  dnamap['C' - 'A'] = ;
    dnamap['G' - 'A'] = ;  dnamap['T' - 'A'] = ;

    char ** ret = malloc(sizeof(char*));
    int curr = ;
    int size = ;
    int key;
    int i = ;

    while (i < )
        key = (key << ) | dnamap[s[i++] - 'A'];
    while (i < len){
        key = ((key << ) & 0xFFFFF) | dnamap[s[i++] - 'A'];
        if (test(map_pattern, key)){
            if (!test(map_exist, key)){
                set(map_exist, key);
                if (curr == size){
                    size *= ;
                    ret = realloc(ret, sizeof(char*)* size);
                }
                ret[curr] = malloc(sizeof(char)* );
                memcpy(ret[curr], &s[i-], );
                ret[curr][] = '\0';
                ++curr;
            }

        }
        else{
            set(map_pattern, key);
        }
    }

    ret = realloc(ret, sizeof(char*)* curr);
    *returnSize = curr;
    return ret;
}

该算法用时 6ms 左右， 非常快