[Swift]LeetCode844. 比较含退格的字符串 | Backspace String Compare

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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?

---

给定 S 和 T 两个字符串，当它们分别被输入到空白的文本编辑器后，判断二者是否相等，并返回结果。 # 代表退格字符。

示例 1：

输入：S = "ab#c", T = "ad#c"
输出：true
解释：S 和 T 都会变成 “ac”。

示例 2：

输入：S = "ab##", T = "c#d#"
输出：true
解释：S 和 T 都会变成 “”。

示例 3：

输入：S = "a##c", T = "#a#c"
输出：true
解释：S 和 T 都会变成 “c”。

示例 4：

输入：S = "a#c", T = "b"
输出：false
解释：S 会变成 “c”，但 T 仍然是 “b”。

提示：

1. 1 <= S.length <= 200
2. 1 <= T.length <= 200
3. S 和 T 只含有小写字母以及字符 '#'。

---

Runtime: 8 ms

Memory Usage: 19.5 MB

class Solution {
    func backspaceCompare(_ S: String, _ T: String) -> Bool 
    {
        var arrS = [Character]()
        var arrT = [Character]()
        for char in S
        {
            if char == "#"
            {
                if arrS.count > 0
                {
                    arrS.removeLast()
                }
            }
            else
            {
                arrS.append(char)
            }
        }
        
        for char in T
        {
            if char == "#"
            {
                if arrT.count > 0
                {
                    arrT.removeLast()
                }
            }
            else
            {
                arrT.append(char)
            }
        }
        return arrS == arrT
    }    
}

---

8ms

class Solution {
    // S2: two pointers space could be 1
    // S1: M+N
    func backspaceCompare(_ S: String, _ T: String) -> Bool {
        return process(S) == process(T)    
    }
    
    private func process(_ S: String) -> String {
        var stack = [Character]()
        for c in Array(S) {
            if c == "#" {
                if stack.count > 0 {
                    stack.removeLast()
                }
            } else {
                stack.append(c)
            }
        }
        return String(stack)
    }
}

---

12ms

class Solution {
    func backspaceCompare(_ S: String, _ T: String) -> Bool {
        var sStack: [Character] = []
        var tStack: [Character] = []

        var sArray = Array(S)
        var tArray = Array(T)
        
        for char in sArray {
            if(char != "#") {
                sStack.append(char)
            } else {
                if(!sStack.isEmpty) {
                    sStack.removeLast()   
                }
            }
        }
        
        for char in tArray {
            if(char != "#") {
                tStack.append(char)
            } else {
                if(!tStack.isEmpty) {
                    tStack.removeLast()   
                }
            }
        }
        
        print(String(sStack))
        print(String(tStack))
        
        if(String(sStack) == String(tStack)) {
            return true
        } else {
            return false
        }
    }
}

---

16ms

class Solution {
    func backspaceCompare(_ S: String, _ T: String) -> Bool {
        var charsS = Array(S), charsT = Array(T)
        
        var i = charsS.count - 1, j = charsT.count - 1
        var countS = 0, countT = 0
        while i >= 0 || j >= 0 {
            while i >= 0 && (countS > 0 || charsS[i] == "#") { 
                if charsS[i] == "#" { 
                    countS += 1
                } else {
                    countS -= 1
                }
                
                i -= 1
            }
            
            while j >= 0 && (countT > 0 || charsT[j] == "#") { 
                if charsT[j] == "#" { 
                    countT += 1
                } else {
                    countT -= 1
                }
                
                j -= 1
            }
            
            if i >= 0 && j >= 0 && charsS[i] == charsT[j] {
                i -= 1
                j -= 1 
            } else {
                break
            }
        }
        return i < 0 && j < 0
    }
}

---

24ms

class Solution {
    func backspaceCompare(_ S: String, _ T: String) -> Bool {
        return backspace(S) == backspace(T)
    }
    
    func backspace(_ string: String) -> String {
        var s = [Character]()

        string.forEach { c in
            if c == "#" {
                if !s.isEmpty {
                    s.removeLast()
                }
            } else {
                s.append(c)
            }
        }
        return String(s)
    }
}

---

24ms

class Solution {
    func backspaceCompare(_ S: String, _ T: String) -> Bool {
        let sRe = reStack(str: S)
        let tRe = reStack(str: T)
        return sRe == tRe;
    }
    
    public func reStack(str: String) -> String {
        var result = [Character]()
        for i in str {
            if (!result.isEmpty) {
                if i != "#" {
                    result.append(i)
                } else {
                    result.removeLast()
                }
            } else {
                if i != "#" {
                    result.append(i)
                }
            }
        }
        return String.init(result);
    }
}