算法小程序 字符串循环右移

字符串循环右移

题目要求

• 编写一个函数，实现循环移动n位字符
• *LoopMove(char pStr, int steps)

题目分析

• 循环右移，需要确定字符串右边有多少位需要依次放到字符串的左边
• 也就是说将字符串分为两部分，通过计算得到两部分的长度
• 动态创建一个字符数组，存储需要移动到左边的字符，然后将原字符串剩下部分拷贝到后边即可

具体代码如下

void LoopMove(char *pStr, int steps)
{
    int len = 0;
    char *temp = pStr;
    char *temp2 = pStr;
    char *temp3 = pStr;
    while(*temp++ != '\0')
    {
        len ++;
    }
    printf("Len:%d\n", len);
    printf("temp2:%s\n", temp2);
    char *temp_str = (char *)malloc((steps%len) * sizeof(char));
    int j = 0;
    for(int i = 0; i < len; i++)
    {
        temp2 ++;
        if(i >= len - (steps % len) - 1)
        {
            strncpy(temp_str, temp2, (steps % len));
            break;
        }
    }
    char *result = (char *)malloc(len * sizeof(char));
    strncpy(result, temp_str, (steps%len));
    for(int i = 0; i < len - (steps % len); i++)
    {
        result[(steps%len) + i] = pStr[i];
    }
    printf("temp_str:%s\n", temp_str);
    printf("result:%s\n", result);
}

整个Demo程序如下：

#include
#include
#include

void LoopMove(char *pStr, int steps)
{
    int len = 0;
    char *temp = pStr;
    char *temp2 = pStr;
    char *temp3 = pStr;
    while(*temp++ != '\0')
    {
        len ++;
    }
    printf("Len:%d\n", len);
    printf("temp2:%s\n", temp2);
    char *temp_str = (char *)malloc((steps%len) * sizeof(char));
    int j = 0;
    for(int i = 0; i < len; i++)
    {
        temp2 ++;
        if(i >= len - (steps % len) - 1)
        {
            strncpy(temp_str, temp2, (steps % len));
            break;
        }
    }
    char *result = (char *)malloc(len * sizeof(char));
    strncpy(result, temp_str, (steps%len));
    for(int i = 0; i < len - (steps % len); i++)
    {
        result[(steps%len) + i] = pStr[i];
    }
    printf("temp_str:%s\n", temp_str);
    printf("result:%s\n", result);
}

int main()
{
    char *pStr = "abcdefghi";
    LoopMove(pStr, 11);
}

Java 的解法

class Solution {
    public String reverseLeftWords(String s, int n) {
        String left = s.substring(0, n);
        String right = s.substring(n);
        return right + left;
    }
}

Done！