題目

We are given that the string "abc" is valid.

From any valid string V, we may split V into two pieces X and Y such that X + Y (X concatenated with Y) is equal to V. (X or Y may be empty.) Then, X + "abc" + Y is also valid.

If for example S = "abc", then examples of valid strings are: "abc", "aabcbc", "abcabc", "abcabcababcc". Examples of invalid strings are: "abccba", "ab", "cababc", "bac".

Return true if and only if the given string S is valid.

Example 1:

Input: "aabcbc"
Output: true
Explanation: 
We start with the valid string "abc".
Then we can insert another "abc" between "a" and "bc", resulting in "a" + "abc" + "bc" which is "aabcbc".

Example 2:

Input: "abcabcababcc"
Output: true
Explanation: 
"abcabcabc" is valid after consecutive insertings of "abc".
Then we can insert "abc" before the last letter, resulting in "abcabcab" + "abc" + "c" which is "abcabcababcc".

Example 3:

Input: "abccba"
Output: false

Example 4:

Input: "cababc"
Output: false

Note:

1. 1 <= S.length <= 20000
2. S[i] is 'a', 'b', or 'c'

題目大意

假設 abc 是有效的字符串，對於任何 字符串 V，如果用 abc 把字符串 V 切成 2 半，X 和 Y，組成 X + abc + Y 的字符串，X + abc + Y 的這個字符串依舊是有效的。X 和 Y 可以是空字符串。

例如，"abc"( "" + "abc" + ""), "aabcbc"( "a" + "abc" + "bc"), "abcabc"( "" + "abc" + "abc"), "abcabcababcc"( "abc" + "abc" + "ababcc"，其中 "ababcc" 也是有效的，"ab" + "abc" + "c") 都是有效的字符串。

"abccba"( "" + "abc" + "cba"，"cba" 不是有效的字符串), "ab"("ab" 也不是有效字符串), "cababc"("c" + "abc" + "bc"，"c"，"bc" 都不是有效字符串), "bac" ("bac" 也不是有效字符串)這些都不是有效的字符串。

任意給一個字符串 S ，要求判斷它是否有效，如果有效則輸出 true。

解題思路

這一題可以類似括號匹配問題，因為 "abc" 這樣的組合就代表是有效的，類似於括號匹配，遇到 "a" 就入棧，當遇到 "b" 字符的時候判斷棧頂是不是 "a"，當遇到 "c" 字符的時候需要判斷棧頂是不是 "a" 和 "b"。最後如果棧都清空了，就輸出 true。�出 true。

參考代碼

package leetcode

func isValid1003(S string) bool {
	if len(S) < 3 {
		return false
	}
	stack := []byte{}
	for i := 0; i < len(S); i++ {
		if S[i] == 'a' {
			stack = append(stack, S[i])
		} else if S[i] == 'b' {
			if len(stack) > 0 && stack[len(stack)-1] == 'a' {
				stack = append(stack, S[i])
			} else {
				return false
			}
		} else {
			if len(stack) > 1 && stack[len(stack)-1] == 'b' && stack[len(stack)-2] == 'a' {
				stack = stack[:len(stack)-2]
			} else {
				return false
			}
		}
	}
	return len(stack) == 0
}