[LeetCode] 844. Backspace String Compare 退格字符串比较

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. 1 <= S.length <= 200
2. 1 <= T.length <= 200
3. S and T only contain lowercase letters and '#' characters.

Follow up:

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

给2个字符串S和T，里面含有#代表退格键，意味着前面的字符会被删除，判断2个字符串是否相等，字符串中只含有小写字母和#。

解法1：用一个栈存字符，循环字符串，如果是字母就加入栈，遇到#而且栈里面有字母就pop出栈里最后的字符。T: O(n), S:(n)

解法2：follow up要求O(N) time and O(1) space，在一个循环里，从后往前处理字符串，用一个变量记录要删除的字符数量，遇到#时变量加1，遇到字符并且变量大于1，变量减1，直到没遇到#并且要变量为0，这时比较两个字符此时是否一样，不一样返回false，如果字符串比较完没有不一样的字符出现，返回ture。

G家：follow up: 如果有大写键CAP

Java: O(1) space

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public boolean backspaceCompare(String S, String T) {         int i = S.length() - 1, j = T.length() - 1;         while (true) {             for (int back = 0; i >= 0 && (back > 0 || S.charAt(i) == '#'); --i)                 back += S.charAt(i) == '#' ? 1 : -1;             for (int back = 0; j >= 0 && (back > 0 || T.charAt(j) == '#'); --j)                 back += T.charAt(j) == '#' ? 1 : -1;             if (i >= 0 && j >= 0 && S.charAt(i) == T.charAt(j)) {                 i--; j--;             } else                 return i == -1 && j == -1;         }     }

Python:

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def backspaceCompare(self, S, T):        i, j = len(S) - 1, len(T) - 1        backS = backT = 0        while True:            while i >= 0 and (backS or S[i] == '#'):                backS += 1 if S[i] == '#' else -1                i -= 1            while j >= 0 and (backT or T[j] == '#'):                backT += 1 if T[j] == '#' else -1                j -= 1            if not (i >= 0 and j >= 0 and S[i] == T[j]):                return i == j == -1            i, j = i - 1, j - 1

Python:

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# Time:  O(m + n) # Space: O(1) import itertools   class Solution(object):     def backspaceCompare(self, S, T):         """         :type S: str         :type T: str         :rtype: bool         """         def findNextChar(S):             skip = 0             for i in reversed(xrange(len(S))):                 if S[i] == '#':                     skip += 1                 elif skip:                     skip -= 1                 else:                     yield S[i]           return all(x == y for x, y in                    itertools.izip_longest(findNextChar(S), findNextChar(T)))

Python: wo O(n) space

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class Solution(object):     def backspaceCompare(self, S, T):         """         :type S: str         :type T: str         :rtype: bool         """         s1, s2 = [], []         for i in range(len(S)):             if S[i] == '#' and s1:                 s1.pop()             elif S[i] == '#':                 continue             else:                 s1.append(S[i])                   for j in range(len(T)):             if T[j] == '#' and s2:                 s2.pop()             elif T[j] == '#':                 continue             else:                 s2.append(T[j])                   return s1 == s2

C++:　　

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bool backspaceCompare(string S, string T) {         int i = S.length() - 1, j = T.length() - 1;         while (1) {             for (int back = 0; i >= 0 && (back || S[i] == '#'); --i)                 back += S[i] == '#' ? 1 : -1;             for (int back = 0; j >= 0 && (back || T[j] == '#'); --j)                 back += T[j] == '#' ? 1 : -1;             if (i >= 0 && j >= 0 && S[i] == T[j])                 i--, j--;             else                 return i == -1 && j == -1;         }     }

　　

　　

 

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