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【待补充】【leetcode_599】【简单】minimum-index-sum-of-two-lists / 两个列表的最小索引总和

拾杨梅记 2022-03-17 阅读 23

文章目录

URL


题目

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分析

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源码

1. 工程结构

├── main_599.c
└── src
    ├── utarray.h
    ├── uthash.h
    ├── utlist.h
    ├── utringbuffer.h
    ├── utstack.h
    └── utstring.h

2.uthash库下载

github:https://github.com/troydhanson/uthash.git
CSDN文档:https://download.csdn.net/download/yujianliam/84995197

3.main

#include <stdio.h> 

#include <limits.h>
#include <string.h>
#include "src/uthash.h"


#define MIN(a,b) ((a)<(b)?(a):(b))

typedef struct {
    char *key;
    int val;
    UT_hash_handle hh;
}HashItem;

void free_hash(HashItem ** obj){
    HashItem *curr = NULL, * next = NULL;
    HASH_ITER(hh , *obj, curr , next){
        HASH_DEL(*obj , curr);
        free(curr);
    }
}

char ** find_restaurant(char ** list1, int list1_size, char ** list2 , int list2_size, int * return_size){
    HashItem * index = NULL;
    HashItem * p_entry = NULL;
    for (int i = 0; i < list1_size; ++i){
        HASH_FIND_STR(index, list1[i], p_entry);
        if (NULL == p_entry){
            p_entry = (HashItem *)malloc(sizeof(HashItem));
            p_entry->key = list1[i];
            p_entry->val = i;
            HASH_ADD_STR(index, key, p_entry);
        }
    }
    char ** ret = (char **)malloc(sizeof(char *) * MIN(list1_size, list2_size));
    int pos = 0;
    int index_sum = INT_MAX;
    for (int i = 0; i < list2_size; ++i){
        HASH_FIND_STR(index, list2[i], p_entry);
        if (NULL != p_entry){
            int j = p_entry->val;
            if (i + j < index_sum){
                pos = 0;
                ret[pos++] = list2[i];
                index_sum = i + j;
            }else if (i + j == index_sum){
                ret[pos++] = list2[i];
            }
        }
    }
    free_hash(&index);
    *return_size = pos;
    return ret;
}

int main()
{
#if 1
    // test data
    const char *data1[] = {"Shogun", "Tapioca Express", "Burger King", "KFC"};
    const char *data2[] = {"Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"};
#else
    const char *data1[] = {"Shogun", "Tapioca Express", "Burger King", "KFC"};
    const char *data2[] = {"KFC", "Shogun", "Burger King"};
#endif
    // define 
    int list1_size = sizeof(data1)/sizeof(data1[1]);
    int list2_size = sizeof(data2)/sizeof(data2[1]);
    char ** list1 = (char **)malloc(sizeof(char *) * list1_size);
    char ** list2 = (char **)malloc(sizeof(char *) * list2_size);
    int ret_size = 0;
    // malloc
    for (int i = 0; i < list1_size; ++i){
        list1[i] = (char *)malloc(sizeof(char) * 50);
    }
    for (int i = 0; i < list2_size; ++i){
        list2[i] = (char *)malloc(sizeof(char) * 50);
    }
    // init
    for (int i = 0; i < list1_size; ++i){
        memset(list1[i], '\0', 50);
    }
    for (int i = 0; i < list2_size; ++i){
        memset(list2[i], '\0', 50);
    }
    // before 
    printf("=> before \n");
    printf("list1 : ");
    for (int i = 0; i < list1_size; ++i){
        strcpy(list1[i], data1[i]);
        printf("%s, ",list1[i]);
    }
    printf("\nlist2 : ");
    for (int i = 0; i < list2_size; ++i){
        strcpy(list2[i], data2[i]);
        printf("%s, ",list2[i]);
    }

    // after
    printf("\n=> after \n");
    char ** ret = find_restaurant(list1, list1_size, list2, list2_size, &ret_size);
    for (int i = 0; i < ret_size; ++i){
        printf("%s , ",ret[i]);
    }
    printf("\n");

    // free
    free(ret);
    for (int i = 0; i < list1_size; ++i){
        free(list1[i]);
    }
    free(list1);
    for (int i = 0; i < list2_size; ++i){
        free(list2[i]);
    }
    free(list2);
    
	return 0;
}

4. LOG参考

=> before
list1 : Shogun, Tapioca Express, Burger King, KFC,
list2 : Piatti, The Grill at Torrey Pines, Hungry Hunter Steakhouse, Shogun,
=> after
Shogun ,

源码概述

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