实操案例：字符串哈希表操作

示例：
输入：
  s = "barfoothefoobarman",
  words = ["foo","bar"]
输出：[0,9]
解释：
从索引 0 和 9 开始的子串分别是 "barfoo" 和 "foobar" 。
输出的顺序不重要, [9,0] 也是有效答案。

unsigned long
hash(unsigned char *str)
{
    unsigned long hash = 5381;

    int c;

    while (c = *str++)

        hash = ((hash << 5) + hash) + c; /* hash * 33 + c */

    return hash;

}

5 -> 110/173
10 -> 143/173
50 -> 170/173
100 -> 170/173
300 -> 172/173
400 -> 173/173

// 字符串最大值，hash表大小，估值和实际数据个数有关

#define MAXWORDCOUNT 1000


static int wordCount[MAXWORDCOUNT];

static int currWordCount[MAXWORDCOUNT];



// ref: http://www.cse.yorku.ca/~oz/hash.html


unsigned long DJBHash(const char* s, int len) {

    unsigned long hash = 5381; // 经验值，hash冲突概率低，分布平衡


    while (len--) {

        hash = (((hash << 5) + hash) + *(s++)) % MAXWORDCOUNT; /* hash * 33 + c */


    }

    return hash;


}



int* findSubstring(char * s, char ** words, int wordsSize, int* returnSize){

    memset(wordCount, 0, sizeof(wordCount));


    *returnSize = 0;



    const int kSLen = strlen(s);

    if (kSLen == 0 || wordsSize == 0) return NULL;



    const int kWordLen = strlen(words[0]);


    // 将单词数量存到哈希表中，key: word, value: 单词数量

    for (int i = 0; i < wordsSize; ++i)


        ++wordCount[DJBHash(words[i], kWordLen)];



    int *result = malloc(sizeof(int) * kSLen);

    for (int i = 0; i < kWordLen; ++i) {


        for (int j = i; j + kWordLen * wordsSize <= kSLen; j += kWordLen) {

            // 统计当前窗口的单词数量


            for (int k = (j == i ? 0 : wordsSize - 1); k < wordsSize; ++k)

                ++currWordCount[DJBHash(s + j + k * kWordLen, kWordLen)];



            // 判断两个哈希表是否相等，即窗口中的单词是否和给定词典完全匹配


            if (memcmp(wordCount, currWordCount, sizeof(wordCount)) == 0)

                result[(*returnSize)++] = j;



            --currWordCount[DJBHash(s + j, kWordLen)];


        }

        // 哈希表清零操作


        memset(currWordCount, 0, sizeof(currWordCount));

    }


    return result;

}