字符串逆序
1、"dog loves pig" to "pig loves dog"问题
image.png
解题步骤：
①翻转整个字符串
②逐个翻转单词
implement by C programmer language：
#include<iostream>
#include<string.h>
using namespace std;
void reverse(char *str, int start, int end);
int main()
{
int flag = 0;
char str[] = "pig loves dog";
reverse(str,0,12);
cout<<str<<endl;
for(int i = 0; i < strlen(str); ++i)
{
if(str[i] == ' ')
{
reverse(str, flag, i - 1);
flag = i + 1;
}
if(i == strlen(str) - 1)
reverse(str, flag, i);
}
cout<<str<<endl;
}
void reverse(char *str, int start, int end)
{
if(start > end) return;
if(start < 0 || end >= strlen(str)) return;
//exchange all elements between start to end
while(start < end)
{
char temp = str[start];
str[start] = str[end];
str[end] = temp;
++start;
--end;
}
}
2、以O(1)的空间复杂度调整字符串
本题的关键在于不能用额外的空间，因此利用字符串逆序来解决此类问题，是非常有效的，也几乎是唯一的解法。
image.png
- 解题步骤
- ①翻转局部字符串
- ②翻转整个字符串
#include<iostream>
#include<string.h>
using namespace std;
void reverse(char *str, int start, int end);
int main()
{
char str[] = "ABCDE";
int k = 2;//给定的移动位置
reverse(str,0,2);
reverse(str,3,4);
reverse(str,0,4);
cout<<str;
}
//逆序start to end之间的字符
void reverse(char *str, int start, int end)
{
if(start > end) return;
if(start < 0 || end >= strlen(str)) return;
//exchange all elements between start to end
while(start < end)
{
char temp = str[start];
str[start] = str[end];
str[end] = temp;
++start;
--end;
}
}
