LeetCode - 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 <= S.length <= 200
1 <= T.length <= 200
S and T only contain lowercase letters and '#' characters.
Follow up:

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

注意，这道题都是需要 time O(n) 和 space O(1)的，所以需要用 two pointer来做，注意最后结束条件：

class Solution {
    public boolean backspaceCompare(String S, String T) {
        if(S == null || T == null){
            return false;
        }
        int i1 = S.length()-1;
        int count1= 0;

        int i2 = T.length()-1;
        int count2 = 0;

        while(i1 >= 0 || i2 >= 0){
            if(i1 >= 0 && S.charAt(i1) == '#'){
                count1++;
                i1--;
            }
            else if (i2 >= 0 && T.charAt(i2) == '#'){
                count2++;
                i2--;
            }
            else if( i1 >= 0 && S.charAt(i1) != '#' && count1 > 0){
                count1--;
                i1--;
            }
            else if(i2 >= 0 && T.charAt(i2) != '#' && count2 > 0){
                count2--;
                i2--;
            }
            else{
                if( i1>=0 && i2 >= 0 && S.charAt(i1) == T.charAt(i2) ){
                    i1--;
                    i2--;
                }
                else{
                    return false;
                }
            }
        }

        return true;

    }
}