30-Day Leetcoding Challenge Day9

两种解法：

1.用stack存储，重新构建字符串，比较是否相等。

2.双指针，有后向前遍历，此解法空间复杂度为O(1)。

JAVA

class Solution {
    public boolean backspaceCompare(String S, String T) {
        return helpback(S).equals(helpback(T));
    }
    public Stack<Character> helpback(String str){
        Stack<Character> stack = new Stack();
        for(int i = 0; i < str.length(); i++){
            if(str.charAt(i) == '#'){
                if(stack.size() != 0)stack.pop();
                continue;
            }
            stack.push(str.charAt(i));
        }
        return stack;
    }
}

class Solution {
    public boolean backspaceCompare(String S, String T) {
        int i = S.length()-1;
        int j = T.length()-1;
        int iskip = 0;
        int jskip = 0;
        while(i >= 0 || j >= 0){
            while(i >= 0){ //!!!找到比较的位置
                if(S.charAt(i) == '#'){
                    iskip++;
                    i--;
                }
                else if(iskip > 0){
                    iskip--;
                    i--;
                }
                else break;
            }
            while(j >= 0){ //!!!找到比较的位置
                if(T.charAt(j) == '#'){
                    jskip++;
                    j--;
                }
                else if(jskip > 0){
                    jskip--;
                    j--;
                }
                else break;
            }
            // If two actual characters are different
            if(i >= 0 && j >= 0 && S.charAt(i) != T.charAt(j))
                return false;
            // If expecting to compare char vs nothing
            if((i >=0 ) != (j >= 0))
                return false;
            i--;
            j--;
        }
        return true;
    }
}

Python3

class Solution:
    def backspaceCompare(self, S: str, T: str) -> bool:     
        return self._back(S) == self._back(T)
    def _back(self, str):
        stack = []
        for i in range(len(str)):
            if str[i] == '#':
                if stack:
                    stack.pop()
                continue
            stack.append(str[i])
        return stack

class Solution:
    def backspaceCompare(self, S: str, T: str) -> bool:     
        i = len(S)-1
        j = len(T)-1
        iskip = 0
        jskip = 0
        while i >= 0 or j >= 0:
            while i >= 0:
                if S[i] == '#':
                    iskip += 1
                    i -= 1
                elif iskip > 0:
                    iskip -= 1
                    i -= 1
                else:
                    break
            while j >= 0:
                if T[j] == '#':
                    jskip += 1
                    j -= 1
                elif jskip > 0:
                    jskip -= 1
                    j -= 1
                else:
                    break
            if i >= 0 and j >= 0 and S[i] != T[j]:
                return False
            if (i >= 0) != (j >= 0):
                return False
            i -= 1
            j -= 1
        return True