C Language: String substitution spaces: Implement a function that replaces each space in a string with "%20".

Ideas:

In the past: replace on the original basis (of course, if the space is sufficient), if you encounter a space after the trip to replace, it will inevitably overwrite the original character, can not be implemented.

Because the space is replaced with "%20", 2 characters are replaced each time, so you can count the total number of empty cells, and then the new array size is "original array size + space number". Processing from the back: encountered non-whitespace, directly moved to the back, encountered a space replaced by "%20." Until the position pointer is inserted and the original array is the pointer coincident position.

#include <stdio.h> #include <string.h>int main () { char arr[] =  "We are  happy. ";  int i = 0; int j = 0; int len = 0; int  Count = 0; len = strlen (arr);  for (i = 0;i < len;i++)  {  if (arr[i] ==  '   ')   {   count ++;    }   } i = len; j = 2 * count + len;  while (i != j && i >= 0)  {  if (arr[i] ==   '   ')   {   arr[j--] =  ' 0 ';   arr[j--] =  ' 2 ';   arr[j--] =  '% ';   i--;  }  else   {   arr[j] = arr[i];   j--;   i--;   }  } lEn = strlen (arr);  for (i = 0; i< len ;i++)  {  printf ( "%c", Arr[i]);  } printf ("\ n");  return 0; }

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C Language: String substitution spaces: Implement a function that replaces each space in a string with "%20".