844. Backspace String Compare

Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.

Example 1:

Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T become "ac".

Example 2:

Input: S = "ab##", T = "c#d#"
Output: true
Explanation: Both S and T become "".

Example 3:

Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".

Example 4:

Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".

Note:

1. 1 <= S.length <= 200
2. 1 <= T.length <= 200
3. S and T only contain lowercase letters and '#' characters.

Follow up:

• Can you solve it in O(N) time and O(1) space?
  

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  题目没看太懂,看完案例看懂了。字符'#'代表退格符,如果在#号前有非#号字符的话就要消去。这样子题目就很浅显易懂了，直接把字符串一个一个对比,如果遇到非#号字符就保存,遇到#号并且#号前有字符的话就消去,最后再比较两个字符串的字符是否相等就行。
  

  void modifyString(char* S) {
      int cur = 0;
      for(int i=0;S[i]!='\0';i++) {
          if(S[i]!='#') {
              S[cur++] = S[i];
          }     
          else if(S[i]=='#' && cur>0) {
              cur--;
          }
      }
      S[cur] = '\0';//因为字符串数组最后的一个字符必须是'\0'
  }
  bool backspaceCompare(char* S, char* T) {
      modifyString(S);
      modifyString(T);
      
      if(strcmp(S,T)==0){//strcmp函数用于比较两个字符串
          return true;
      }else{
          return false;
      }
  }

   

   

   

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