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Description
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 <= S.length <= 200
- 1 <= T.length <= 200
- S and T only contain lowercase letters and ‘#’ characters.
Follow up:
Can you solve it in O(N) time and O(1) space?
分析
题目的意思为:判断两个字符串最终的输出是否是一样的,其中#代表键盘回退/删除
- 双指针法,从后往前遍历,代码写了三个循环,很容易懂
代码
class Solution {
public:
bool backspaceCompare(string S, string T) {
int i=S.size()-1;
int j=T.size()-1;
int skipS=0;
int 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&&S[i]!=T[j]){
return false;
}
if((i>=0)!=(j>=0)){
return false;
}
i--;
j--;
}
return true;
}
};