[LeetCode] 1003. Check If Word Is Valid After Substitutions

轉自LeetCode

Given a string s, determine if it is valid.

A string s is valid if, starting with an empty string t = "", you can transform t into s after performing the following operation any number of times:

• Insert string "abc" into any position in t. More formally, t becomes tleft + "abc" + tright, where t == tleft + tright. Note that tleft and tright may be empty.

Return true if s is a valid string, otherwise, return false.

 

Example 1:

Input: s = "aabcbc"
Output: true
Explanation:
"" -> "abc" -> "aabcbc"
Thus, "aabcbc" is valid.

Example 2:

Input: s = "abcabcababcc"
Output: true
Explanation:
"" -> "abc" -> "abcabc" -> "abcabcabc" -> "abcabcababcc"
Thus, "abcabcababcc" is valid.

Example 3:

Input: s = "abccba"
Output: false
Explanation: It is impossible to get "abccba" using the operation.

Example 4:

Input: s = "cababc"
Output: false
Explanation: It is impossible to get "cababc" using the operation.

 

Constraints:

• 1 <= s.length <= 2 * 104
• s consists of letters 'a', 'b', and 'c'

Solution

這題可以看作是 20. Valid Parentheses 的衍生題

想法稍微變一下，從原本的左括號右括號配對，變成 abc 配對

意思是，c 的前面，一定要有 ab，不然就是不合法

所以，當遇到 c 的時候，去檢查前面是否有b和a

如果沒有，就回傳 false，有就繼續檢查下去

最後檢查 stack 裡面是否都用光了

kotlin

	class Solution {
	fun isValid(s: String): Boolean {
	val stack = mutableListOf<Char>()
	for(c in s) {
	if(c == 'c') {
	if(stack.isEmpty() || stack.removeAt(stack.lastIndex) != 'b') {
	return false
	}
	if(stack.isEmpty() || stack.removeAt(stack.lastIndex) != 'a') {
	return false
	}
	} else {
	stack.add(c)
	}
	}
	return stack.isEmpty()
	}
	}

view raw check_if_word_is_valid_after_substitutions.kt hosted with ❤ by GitHub