[LeetCode] 844. Backspace String Compare

轉自LeetCode

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?

<Solution>

想法如下

• 用 stack 來組出最後的字串，然後再比較兩個最後的字串是不是一樣

code 如下

Java

	class Solution {
	public boolean backspaceCompare(String S, String T) {
	return build(S).equals(build(T));
	}
	private String build(final String str) {
	Stack<Character> stack = new Stack<>();
	for(char c : str.toCharArray()) {
	if(c != '#') {
	stack.push(c);
	}
	else if(!stack.empty()) {
	stack.pop();
	}
	}
	return String.valueOf(stack);
	}
	}

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