844. Compare strings with backspace
Given Sand Ttwo strings, when they are inputted to a blank text editor, it is determined whether the two are equal, and returns the result. #Represents the backspace character.
Note: If you enter a backspace character for empty text, the text will remain empty.
Example 1:
输入：S = "ab#c", T = "ad#c"
输出：true
解释：S 和 T 都会变成 “ac”。
Example 2:
输入：S = "ab##", T = "c#d#"
输出：true
解释：S 和 T 都会变成 “”。
Example 3:
输入：S = "a##c", T = "#a#c"
输出：true
解释：S 和 T 都会变成 “c”。
Example 4:
输入：S = "a#c", T = "b"
输出：false
解释：S 会变成 “c”，但 T 仍然是 “b”。
prompt:
- 1 <= S.length <= 200
- 1 <= T.length <= 200
- S and T only contain lowercase letters and the character'#'.
Advanced:
You can use the O(N)time complexity and O(1)space complexity of solving this problem?
Method 1: Refactor the string
bool backspaceCompare(char* S, char* T) {
return strcmp(build(S), build(T)) == 0;
}
char* build(char* str) {
int n = strlen(str), len = 0;
char* ret = malloc(sizeof(char) * (n + 1));
for (int i = 0; i < n; i++) {
if (str[i] != '#') {
ret[len++] = str[i];
} else if (len > 0) {
len--;
}
}
ret[len] = '\0';
return ret;
}
Method 2: Double pointer
Whether a character is deleted or not depends only on whether there is something behind the character #, so we can traverse from back to front. In this way, it can be immediately determined whether the character is deleted.
bool backspaceCompare(char* S, char* T) {
int i = strlen(S) - 1, j = strlen(T) - 1;
int skipS = 0, skipT = 0;
while (i >= 0 || j >= 0) {
while (i >= 0) {
if (S[i] == '#') {
skipS++, i--;
} else if (skipS > 0) {
skipS--, i--;
} else {
break;
}
}
while (j >= 0) {
if (T[j] == '#') {
skipT++, j--;
} else if (skipT > 0) {
skipT--, j--;
} else {
break;
}
}
if (i >= 0 && j >= 0) {
if (S[i] != T[j]) {
return false;
}
} else {
if (i >= 0 || j >= 0) {
return false;
}
}
i--, j--;
}
return true;
}