个人学习编程(3-23) leetcode刷题

字符串的移动：

您将得到一个包含小写英文字母的字符串s和一个矩阵shift，其中

shift[i] = [direction， amount]:

direction可以为0（左移）或1（右移）。
amount是字符串s被移动的量。
左移1表示删除s的第一个字符并将其附加到末尾。
类似地，向右移动1意味着删除最后一个字符并将其添加到开头。

两层for循环：

char * stringShift(char * s,int** shift,int shiftSize,int* shiftColSize){
    for (int i = 0; i < shiftSize; i++){
        
        int len = strlen(s);
        int direction = shift[i][0];//二维数组中的元素的值第一位是方向
        int amount = shift[i][1];//第二位是代表次数
        if (direction == 0){ //左移
            for (int j = 0; j < amount; j++){
            int temp = s[0];
                for (int k = 0; k < len -1; k++){
                    s[k] = s[k + 1];
                }
            s[len - 1] = temp;
        }
    }
        else{
            for (int r = 0; r < amount; r++){
                int temp = s[len - 1];
                    for (int c = len - 1; c > 0; c--){
                        s[c] = s[c - 1];
                    }
            s[0] = temp;
            }
        }
    }
    return s;
}

左移1 == 右移len - 1：

char * stringShift(char * s,int** shift,int shiftSize,int* shiftColSize){
    int len = strlen(s);
    
    int direction = shift[i][0];
    int amount = shift[i][1];

    amount = amount % len;

    if (direction == 0){
        amount = amount;
    }else{
        amount = len - amount;
    

    for (int k = 1; k < amount; k++){
        char temp = s[0];
        for (int j = 0; j < len - 1; j++){
            s[j] = s[j + 1];
        }
        s[len - 1] = temp;
        }
    }
    return s;
}

Finally （写翻转函数）：

void reverse(char* s,int start,int end){
    end--;
    char temp;
    while (start < end){
        temp = s[start];
        s[start] = s[end];
        s[end] = temp;
        start++;
        end--;
    }
}

char* stringShift(char * s,int** shift,int shiftSize,int* shiftColSize){
    int len = strlen(s);
    int totalAmount = 0;
    for (int i = 0; i < shiftSize; i++){
        int direction = shift[i][0];
        int amount = shift[i][1];

        amount = amount % len;
        if (direction == 1){
            amount = len - amount;
        }
        totalAmount = totalAmount + amount; 
    }
    totalAmount %= len;
    
    reverse(s, 0, totalAmount);
    reverse(s,totalAmount,len);
    reverse(s,0,len);
    return s;
}

数组除了自己的乘积：

输入：

Copy Code

4 1 2 3 4

输出：

Copy Code

24 12 8 6

遍历：

 这个方法很好，就是算出来全部的乘积，然后除以每一个重新赋值到这个位置就可以得到任务需求

#include <stdio.h>

long long multiply(int* arr, int n) {
    long long total_product = 1;
    for (int i = 0; i < n; i++) {
        total_product *= arr[i];
    }
    return total_product;
}

int main() {
    int n;
    scanf("%d", &n);
    int arr[n];

    // 输入数组
    for (int i = 0; i < n; i++) {
        scanf("%d", &arr[i]);
    }

    // 计算整个数组的乘积
    long long total_product = multiply(arr, n);

    // 输出每个位置的乘积（总乘积除以当前元素）
    for (int i = 0; i < n; i++) {
        arr[i] = total_product / arr[i];
        printf("%lld ", arr[i]);  // 输出 long long 类型
    }
    printf("\n");  // 输出完毕后换行

    return 0;
}

 按照0的数量：

#include <bits/stdc++.h>


int* productExceptSelf(int* nums,int numsSize,int* returnSize){
    *returnSize = numsSize;

    int* result = (int *)malloc(sizeof(int)*numsSize);
    int numberOfZeros = 0;
    for (int i = 0; i < numsSize; i++){
        if (nums[i] == 0){
            numberOfZeros++;
        }
    }

    if (numberOfZeros == 1){
        int total = 1;
        for (int i = 0; i < numsSize; i++){
            if (nums[i] != 0){
                total *= nums[i];
            }
        }
        
        for (int i = 0; i < numsSize; i++){
            if (nums[i] != 0){
                result[i] = 0;
            } else{
                result[i] = total;
            }
        }
    }
    
    else if (numberOfZeros >= 2){
        for (int i = 0; i < numsSize; i++){
            result[i] = 0;
        } 
    }
    else{
        int total = 1;
        for (int k = 0; k < numsSize; k++){
            total *=nums[k];
        }
        for (int i = 0; i < numsSize; i++){
            result[i] = total / nums[i];
        }
    }
    return result;

}

待理解的最优解：

#include <bits/stdc++.h>
int* productExceptSelf(int* nums,int numsSize,int* returnSize){
    *returnSize = numsSize;
    int* result = (int*)malloc(sizeof(int)*numsSize);

    
    int rights[numsSize];
    rights[numsSize - 1] = 1;
    for(int i = numsSize - 2;i >= 0;i--){
        rights[i] = rights[i + 1] * nums[i + 1];
    }

    int left = 1;
    for (int i = 0; i < numsSize; i++){
        result[i] = left * rights[i];
        left = left * nums[i];
    }
    return result;
    
}

X进制转为Y进制：（10进制作为中转站）

#include <stdio.h>
#include <bits/stdc++.h>
#include <stdlib.h>

int main() {
    char s1[105];
    int len,x1,x2;
    int count = 0;
    char s2[105];
    int i;
    int ans = 0;
    printf("请输入进制字符串以及要转的进制");
    scanf("%s %d %d",&s1,&x1,&x2);
    len = strlen(s1);
    for (i = 0; i < len; i++){
        ans = ans * x1;
        if(s1[i] >= '0' && s1[i] <= '9'){
            ans += s1[i] -'0';
        }else{
            ans += s1[i] - 'A' + 10;
        }
    }
    printf("这个数转为10进制是：%d\n",ans);

    while (ans > 0){
        int w = ans % x2;
        if(w < 10){
            s2[count++] = w + '0';
        }else{
            s2[count++] = (w-10) + 'a';
        }
        ans/=x2;
    }
    for (int i = count - 1; i >= 0; i--){
        printf("%c",s2[i]);
    }
    return 0;
}