Backspace String Compare
Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character.
Note that after backspacing an empty text, the text will continue empty.
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 = "b"
Output: false
Explanation: s becomes "c" while t becomes "b".
Constraints:
1 <= s.length, 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?
思路一：用栈实现
public boolean backspaceCompare(String s, String t) {
char[] chars1 = s.toCharArray();
char[] chars2 = t.toCharArray();
Stack<Character> stack1 = new Stack<>();
for (char c : chars1) {
if (c == '#' && !stack1.isEmpty()) {
stack1.pop();
} else if (c != '#') {
stack1.push(c);
}
}
Stack<Character> stack2 = new Stack<>();
for (char c : chars2) {
if (c == '#' && !stack2.isEmpty()) {
stack2.pop();
} else if (c != '#') {
stack2.push(c);
}
}
return stack1.equals(stack2);
}
思路二： 从字符串后面往前面遍历，可以省去栈的调用
public boolean backspaceCompare2(String s, String t) {
char[] chars1 = s.toCharArray();
char[] chars2 = t.toCharArray();
StringBuilder s1 = new StringBuilder();
int count = 0;
for (int i = chars1.length - 1; i >= 0; i--) {
char c = chars1[i];
if (c == '#') {
count++;
} else {
if (count == 0) {
s1.append(c);
} else {
count--;
}
}
}
StringBuilder s2 = new StringBuilder();
count = 0;
for (int i = chars2.length - 1; i >= 0; i--) {
char c = chars2[i];
if (c == '#') {
count++;
} else {
if (count == 0) {
s2.append(c);
} else {
count--;
}
}
}
return s1.toString().equals(s2.toString());
}