Preface

        When it comes to string reverse order, we usually think of , Define two variables that point to the beginning and end of the array left and right, stay left<rigth Under the premise of continuous exchange of array contents . And today we see that the reverse order difficulty of the string will be even higher .

One 、 Topic analysis

        Assuming a string "i like beijing." After normal reverse order is ".gnijieb ekil i" The result given in the title is "beijing. like i.

Here is an idea to reverse the normal order of strings to ".gnijieb ekil i", Then reverse each word again to get

To the desired result .

Two 、 Code implementation

#include<stdio.h>
#include<string.h>
void reverse(char* left,char* right) {
	while (left<right) {
		char temp = *left;
		*left = *right;
		*right = temp;
		left++;
		right--;
	}
}
int main() {
	// Input 
	char arr[101] = {0};
	gets_s(arr);// because scanf When a space is encountered, it will stop reading , So using gets Read in 
	// The reverse 
	// The whole is in reverse order 
	int len = strlen(arr);
	reverse(arr,arr+len-1 );
	// Single reverse order 
	char* start = arr;// Define the start position and end position 

	while (*start) {
		char* end = start;

		while (*end != ' ' && *end!='\0') {//*end representative end!='\0'
			end++;
		}
		reverse(start, end - 1);// Single reverse order 
		if (*end!='\0') {
			end++;
		}
		start = end;
		
	}

	// Output 
	printf("%s\n", arr);

	return 0;
}

summary

          This problem has certain requirements for the knowledge of mutual calls between pointers and string arrays , If you are not familiar with array knowledge, you can only implement it at the array level . If there are deficiencies, please call home for more advice .