leetode.599. Minimum Index Sum of Two Lists

Question

Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings. 

You need to help them find out their common interest with the least list index sum. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.

Example 1:

Input:["Shogun", "Tapioca Express", "Burger King", "KFC"]["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]Output: ["Shogun"]Explanation: The only restaurant they both like is "Shogun".

Example 2:

Input:["Shogun", "Tapioca Express", "Burger King", "KFC"]["KFC", "Shogun", "Burger King"]Output: ["Shogun"]Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).

Note:

1. The length of both lists will be in the range of [1, 1000].
2. The length of strings in both lists will be in the range of [1, 30].
3. The index is starting from 0 to the list length minus 1.
4. No duplicates in both lists.

Solution

/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */char** findRestaurant(char** list1, int list1Size, char** list2, int list2Size, int* returnSize) {        int i,j,sum=list1Size+list2Size-2,index=0;    int n = list1Size > list2Size ? list1Size : list2Size;    char **str = (char**)malloc(n * sizeof(char *));        for (i = 0; i < list1Size; i++) {        for (j = 0; j < list2Size; j++) {            if (strcmp(list1[i], list2[j]) == 0) {                if (i+j <= sum) {                    sum = i + j;                    *(str+index) = (char)malloc(sizeof(char));                    *(str+index) = list1[i];                    index++;                }                break;            }        }    }        *returnSize = index;        return str;}

Solution with hash map (别人的最优解)

/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */#include <string.h>#define HASHKEY 0xFF#define NAMESIZEALIGN8 0x20#define DEBUG 0char** findRestaurant(char** list1, int list1Size, char** list2, int list2Size, int* returnSize) {    typedef struct hash_struct {        char *name;        int index;        struct hash_struct *next;    }hashs;        int hindex, i, j, anssum = list1Size + list2Size, max;    hashs *hashmap[HASHKEY + 1], *hcur;    char ans[list1Size + list2Size][31], *acur, **ans_test;        for (i = 0; i < HASHKEY + 1; i++)        hashmap[i] = NULL;        for (i = 0; i < list1Size; i++) {        hindex = (list1[i][0] + list1[i][1]) & HASHKEY;        hcur = malloc(sizeof(hashs));        hcur->name = list1[i];        hcur->index = i;        if (hashmap[hindex])            hcur->next = hashmap[hindex];        else            hcur->next = NULL;        hashmap[hindex] = hcur;    }    for (i = 0; i < list2Size; i++) {        hindex = (list2[i][0] + list2[i][1]) & HASHKEY;        hcur = hashmap[hindex];        while (hcur) {            if (strcmp(list2[i], hcur->name))                hcur = hcur->next;            else {                if (hcur->index + i < anssum) {                    anssum = hcur->index + i;                    *returnSize = 1;                    strcpy(ans[0], list2[i]);                } else if (hcur->index + i == anssum) {                    j = *returnSize;                    strcpy(ans[j], list2[i]);                    (*returnSize)++;                }                break;            }        }    }        if (*returnSize) {        max = *returnSize;        ans_test = malloc(max * sizeof(char *));        for (i = 0; i < max; i++) {            ans_test[i] = malloc(NAMESIZEALIGN8 * sizeof(char));            strcpy(ans_test[i], ans[i]);        }        return ans_test;    }    return NULL;}