Java 实现 classSolution{publicbooleanbackspaceCompare(String S, String T){inti=S.length() -1, j = T.length() -1;intsSkip=0, tSkip =0;while(i >=0|| j >=0) {// 找到字符串S所对应的结果的下一个字符while(i >=0) {if(S.charAt(i) =
链接:https://leetcode-cn.com/problems/backspace-string-compare python # 比较退格字符串 classSolution: defbackspaceStringCompare1(self,S:str,T:str)->bool: """ 栈的思想, 时间O(m+n),空间借助栈O(m+n) :param S: :param T: :return: """ defbuildNewString(s:str)->str: ret = list(...
Can you solve this real interview question? Backspace String Compare - Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character. Note that after backspacing an empty text, the
先用一种最容易想到的方法实现——直接拼接出最终字符串。 StringBuilder之间的比较不能直接用equals,需要先转换为String再用equals。 classSolution{ publicbooleanbackspaceCompare(Strings,Stringt) { StringBuilderS=newStringBuilder(),T=newStringBuilder(); for(inti=0;i<s.length();i++) { if(s.charAt(i)=...
1 class Solution { 2 public boolean backspaceCompare(String S, String T) { 3 return build(S).equals(build(T)); 4 } 5 6 public String build(String S) { 7 Stack<Character> stack = new Stack(); 8 for (char c : S.toCharArray()) { ...
master LeetCode-Solution/[844]比较含退格的字符串BackspaceCompare.swift Go to file 20 lines (18 sloc) 500 Bytes Raw Blame class Solution {func backspaceCompare(_ S: String, _ T: String) -> Bool { return sortOutStr(str: S).elementsEqual(sortOutStr(str: T)) }...
class Solution { public boolean backspaceCompare(String S, String T) { int i = S.length() - 1, j = T.length() - 1; int sSkip = 0, tSkip = 0; while (i >= 0 || j >= 0) { // 找到字符串S所对应的结果的下一个字符 while (i >= 0) { if (S.charAt(i) == '#') ...
[LeetCode] 844. Backspace String Compare 退格字符串比较 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...
classSolution {publicbooleanbackspaceCompare(String S, String T) {if(S ==null|| T ==null){returnfalse; }inti1 = S.length()-1;intcount1= 0;inti2 = T.length()-1;intcount2 = 0;while(i1 >= 0 || i2 >= 0){if(i1 >= 0 && S.charAt(i1) == '#'){ ...
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