栈stack
844. Backspace String Compare Easy
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”.
Note:
1 <= S.length <= 200 1 <= T.length <= 200 S and T only contain lowercase letters and ‘#’ characters.
这个比较好理解了，对比打字输出的内容是否一致。 使用#代表退格键Backspace。
使用stack还是比较好理解，按照正常的思路就可以解出来。但写代码还是出现了bug，不够仔细。
变量两个String，将每个String最后剩下的内容存在stack中，然后对比两个stack即可。 注意边界条件。
Runtime: 2 ms, faster than 50.07% of Java online submissions for Backspace String Compare. Memory Usage: 34.4 MB, less than 100.00% of Java online submissions for Backspace String Compare.
public boolean backspaceCompare(String S, String T) {
Stack<Character> s = new Stack<>(), t = new Stack<>();
for (int i = 0; i < S.length(); i++) {
char c = S.charAt(i);
if(c == '#') {
if(!s.isEmpty())
s.pop();
}else {
s.push(c);
}
}
for (int i = 0; i < T.length(); i++) {
char c = T.charAt(i);
if(c == '#') {
if(!t.isEmpty())
t.pop();
}else {
t.push(c);
}
}
if(s.isEmpty() && t.isEmpty()) {
return true;
}
if(s.size() != t.size()) {
return false;
}else {
while (!s.isEmpty() && !t.isEmpty() && s.peek() == t.peek()) {
s.pop();
t.pop();
}
return s.isEmpty() && t.isEmpty();
}
}