题目
Given two strings S and T, return 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 = "#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".
方法1：使用双栈。
既然回车就是删除一个字符，那非常符合进栈出栈的思路。没遇到一个#就是出栈一个字符（如果栈空则不管）。以此为思路，代码如下,：
public boolean backspaceCompare2(String S, String T) {
Stack<Character> s1 = new Stack<>();
Stack<Character> t1 = new Stack<>();
for (int i = 0; i < S.length(); i++) {
if (S.charAt(i) == '#') {
if (!s1.empty()) {
s1.pop();
}
} else {
s1.push(S.charAt(i));
}
}
for (int i = 0; i < T.length(); i++) {
if (T.charAt(i) == '#') {
if (!t1.empty()) {
t1.pop();
}
} else {
t1.push(T.charAt(i));
}
}
if (s1.size() != t1.size()) {
return false;
} else {
while (!s1.empty()) {
Character pop = s1.pop();
Character pop1 = t1.pop();
if (pop != pop1) {
return false;
}
}
}
return true;
}
此方法的时间复杂度O（M+N）， M和N分别是S和T的长度， 空间复杂度为O(M+N)， 能否降低空间复杂度？
引入双指针解法。
方法2 双指针
每当遇到一个#，代表的前面一个字符可以回退，
因此分贝将2个字符串从后向前遍历，遇到一个#则回退一个字符，可以记录有多少个#，代表可以回退的字符个数。
public boolean backspaceCompare(String S, String T) {
//
int s1Len = S.length() - 1;
int tLen = T.length() - 1;
int skipS = 0;
int skipT = 0;
while (s1Len >= 0 || tLen >= 0) {
while (s1Len >= 0) {
if (S.charAt(s1Len) == '#') {
skipS++;
s1Len--;
} else if (skipS > 0) {
skipS--;
s1Len--;
} else {
break;
}
}
while (tLen >= 0) {
if (T.charAt(tLen) == '#') {
skipT++;
tLen--;
} else if (skipT > 0) {
skipT--;
tLen--;
} else {
break;
}
}
if (s1Len >= 0 && tLen >= 0 && S.charAt(s1Len) != T.charAt(tLen)) {
return false;
}
if ((s1Len >=0 && tLen < 0) || (s1Len <0 && tLen >= 0) ) {
return false;
}
s1Len--;
tLen--;
}
return true;
}
最后代码如下：
//Given two strings S and T, return if they are equal when both are typed into e
//mpty 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 = "#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?
//
//
//
//
//
// Related Topics Two Pointers Stack
// 👍 1935 👎 97
package leetcode.editor.en;
//Java：Backspace String Compare
import java.util.Stack;
public class P844BackspaceStringCompare {
public static void main(String[] args) {
Solution solution = new P844BackspaceStringCompare().new Solution();
String s = "a#c";
String t = "b";
System.out.println(solution.backspaceCompare(s, t));
System.out.println("____");
s = "a##c";
t = "#a#c";
System.out.println(solution.backspaceCompare(s, t));
System.out.println("____");
// "xywrrmp" "xywrrm#p
s = "xywrrmp";
t = "xywrrm#p";
System.out.println(solution.backspaceCompare(s, t));
System.out.println("____");
// "bxj##tw" "bxj###tw"
s = "bxj##tw";
t = "bxj###tw";
System.out.println(solution.backspaceCompare(s, t));
}
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
/**
* 双指针
* @param S
* @param T
* @return
*/
public boolean backspaceCompare(String S, String T) {
//
int s1Len = S.length() - 1;
int tLen = T.length() - 1;
int skipS = 0;
int skipT = 0;
while (s1Len >= 0 || tLen >= 0) {
while (s1Len >= 0) {
if (S.charAt(s1Len) == '#') {
skipS++;
s1Len--;
} else if (skipS > 0) {
skipS--;
s1Len--;
} else {
break;
}
}
while (tLen >= 0) {
if (T.charAt(tLen) == '#') {
skipT++;
tLen--;
} else if (skipT > 0) {
skipT--;
tLen--;
} else {
break;
}
}
if (s1Len >= 0 && tLen >= 0 && S.charAt(s1Len) != T.charAt(tLen)) {
return false;
}
if ((s1Len >=0 && tLen < 0) || (s1Len <0 && tLen >= 0) ) {
return false;
}
s1Len--;
tLen--;
}
return true;
}
/**
* 栈方法
* @param S
* @param T
* @return
*/
public boolean backspaceCompare2(String S, String T) {
Stack<Character> s1 = new Stack<>();
Stack<Character> t1 = new Stack<>();
for (int i = 0; i < S.length(); i++) {
if (S.charAt(i) == '#') {
if (!s1.empty()) {
s1.pop();
}
} else {
s1.push(S.charAt(i));
}
}
for (int i = 0; i < T.length(); i++) {
if (T.charAt(i) == '#') {
if (!t1.empty()) {
t1.pop();
}
} else {
t1.push(T.charAt(i));
}
}
if (s1.size() != t1.size()) {
return false;
} else {
while (!s1.empty()) {
Character pop = s1.pop();
Character pop1 = t1.pop();
if (pop != pop1) {
return false;
}
}
}
return true;
}
}
//leetcode submit region end(Prohibit modification and deletion)
}
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