如何将数组（或 c 字符串）的元素向左移动给定的数字索引
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/26745018/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
how to shift elements of array (or c-string) left by a given number indexes
提问by
I am writing a function to shift the characters of my c-string left by a given number of characters. Currently the function will shift the characters left but I am losing one. I know it is some sort of indexing issue with my for loop, but I can't pin it down.
我正在编写一个函数来将我的 c 字符串的字符向左移动给定数量的字符。目前该功能将向左移动字符，但我丢失了一个。我知道这是我的 for 循环的某种索引问题，但我无法确定。
EDIT: By shift left I mean:
编辑：左移我的意思是：
Given a starting c-string of a, b, c, d if shifted left one index this same array would equal b, c, d, a if shifted left two indexes this same c-string would equal c, d, a, b
给定 a, b, c, d 的起始 c-string 如果左移一个索引，这个相同的数组将等于 b, c, d, a 如果左移两个索引，这个相同的 c-string 将等于 c, d, a, b
Here is my code so far:
到目前为止，这是我的代码：
#include <iostream>
using namespace std;
void shiftleft (char myarray[], int size, int shiftBy)
{
char temp;
for (int i=size-1; i > 0; i--)
{
temp = myarray[size+shiftBy];
myarray[size+shiftBy] = myarray[i];
myarray[i] = temp;
}
}
int main() {
char myarray[20] = "test";
int size = 4;
shiftleft (myarray, size, 1);
for(int i = 0; i < size+1; i++){
cout << myarray[i];
}
return 0;
}
Here is my working function that shifts each element to the right, all I need to do is reverse this loop, and move the elements left, as in this way: <----
这是我将每个元素向右移动的工作函数，我需要做的就是反转这个循环，并将元素向左移动，如下所示：<----
//function bloack
void shiftright (char myarray[], int size, int shiftBy)
{
if(shiftBy > size){
shiftBy = shiftBy - size;
}
if(size == 1){
//do nothing
}
else{
char temp;
//for loop to print the array with indexes moved up (to the right) --> by 2
for (int i=0; i < size; i++)
{
temp = myarray[size-shiftBy];
myarray[size-shiftBy] = myarray[i];
myarray[i] = temp;
}
}
}
采纳答案by
With a little fanangling, I was able to get it to work. Here is my functioning function :)
通过一点点fanangling，我能够让它工作。这是我的功能:)
The issue was that I needed to assign element i to i+shiftBy, and only repeat the loop while i < size-shiftBy.
问题是我需要将元素 i 分配给 i+shiftBy，并且只在 i < size-shiftBy 时重复循环。
//function bloack
void shiftLeft (char myarray[], int size, int shiftBy)
{
if(shiftBy > size){
shiftBy = shiftBy - size;
}
if(size == 1){
//do nothing
}
else{
char temp;
//for loop to print the array with indexes moved up (to the left) <-- by 2
for (int i=0; i < size-shiftBy; i++)
{//EXAMPLE shift by 3 for a c-string of 5
temp = myarray[i];//temp = myarray[0]
myarray[i] = myarray[i + shiftBy];//myarray[0] == myarray[2]
myarray[i + shiftBy] = temp;//myarray[2] = temp(value previously at index i)
}
}
}
回答by morteza khadem
if you want shift string to left and don't rotate, you can use this code:
如果您想将字符串向左移动并且不旋转，您可以使用以下代码：
void shiftLeft (char *string,int shiftLength)
{
int i,size=strlen(string);
if(shiftLength >= size){
memset(string,'char* name = strdup("Hello");
char* str_ptr = (char*) malloc(sizeof(char));
strcpy(str_ptr, name);
int j = 3 // len of shift
for(int i = 0; i < strlen(str_ptr); i++){
printf("%c", str_ptr[(i+j)%strlen(str_ptr)]);
}
',size);
return;
}
for (i=0; i < size-shiftLength; i++){
string[i] = string[i + shiftLength];
string[i + shiftLength] = '##代码##';
}
}
回答by athyagat
// Suppose if j is your len of shift
// 假设 j 是你的移位长度
##代码##