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".
分析：题目要求很简单，两个字符串S和T，每个字符串中都有'#'字符，遇到'#'字符就要往后删掉一个，最后判断操作后的结果是否相等。题目很好理解，关键要能想到'#'出现的位置，如果#前面没有字符了，那就应该忽略。找准这个，题目的解法就比较容易了。这里我用了一个helper函数来完成对字符串的back操作。用一个stringbuffer来记录结果。代码如下：
1 public boolean backspaceCompare(String S, String T) { 2 return helper(S).equals(helper(T)); 3 } 4 private String helper(String s){ 5 StringBuilder res = new StringBuilder(""); 6 int cur = 0; 7 while ( cur < s.length() ){ 8 char c = s.charAt(cur); 9 if ( c == '#' && res.length() != 0) res.deleteCharAt(res.length()-1); 10 if ( c != '#') res.append(c); 11 cur++; 12 } 13 // System.out.println(res); 14 return res.toString(); 15 }
提交结果还是比较好的，运行时间5ms，击败了99.84的提交次数。