[算法刷题]-二分法

这是我参与更文挑战的第5天，活动详情查看：更文挑战

好可惜啊，昨天不可抗力的缺席了一天，想要的掘金徽章套装盒没了，看来无缘徽章了😢😢😢~

开始算法刷题！！！

入门练手

数组

数组是存放在连续内存空间上的相同类型数据的集合。

二分法

编号35：搜索插入位置

给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。

class Solution{
public:
    int searchInsert(vector<int>& nums, int target){
        for(int i = 0; i< nums.size();i++){
            
        }
    }
};

shuffleing Machine

A deck of 54 cards according tp a given random order and repeats for a given number of times.It is assumed that the initial status of a card deck is in the following order:

S1,S2……,S13,H1,……,C1,……,D1,……,D13,J1,J2

where S stands for Spade,H for Heart C for Club, D for Diamond, and D for Joker.

A given order is a permutation of distinct integers in [1, 54].

If the number at the i-th position is j, it means to move the card from position i to position j.

For example, suppose we only have 5 cards: S3, H5, C1, D13 and J2.

Given a shuffling order {4, 2, 5, 3, 1}, the result will be: J2, H5, D13, S3, C1. If we are to repeat the shuffling again, the result will be: C1, H5, S3, J2, D13.

#include <stdio.h>

int main() {
    int cards[55];
    int mids[55];
    int ends[55];
    int times = 0;
    for (int i = 1; i < 55; i++) {
        cards[i] = i;
    }
    scanf("%d", &times);
    for (int i = 1; i < 55; i++)
    {
        scanf("%d", &mids[i]);
    }
    for (int j = 0; j < times; j++) {
        for (int i = 1; i < 55; i++) {
            ends[mids[i]] = cards[i];
        }
        for (int a = 1; a < 55; a++) {
            cards[a] = ends[a];
        }
    }
    for (int i = 1; i < 55; i++) {
        int temp = cards[i];
        if (temp <= 13) printf("S%d", ends[i]);
        else if (temp < 26) printf("H%d", temp%13);
        else if (temp == 26) printf("H13");
        else if(temp < 39) printf("C%d",temp%13);
        else if(temp == 39) printf("C13");
        else if (temp < 52) printf("D%d", temp%13);
        else if (temp == 52) printf("D13");
        else printf("J%d", temp%13);
        if (i != 54) printf(" ");
    }
    return 0;
}

思考和总结

改进：最后的输出过于繁琐，可以简化如下。

for(int i = 1;i<=55;i++)
{
    if(i != 1) printf(" ");
    cards[i]--;
    printf("%c%d",CHA[cards[i]/13],cards[i]%13+1);//通过字符数组输出字母，通过除法判断字母；后者通过原数字-1避开13%13==0的情况
}