1003. Check If Word Is Valid After Substitutions #
题目 #
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 <= S.length <= 20000
- S[i] is ‘a’, ‘b’, or ‘c’
题目大意 #
假设 abc 是有效的字符串，对于任何 字符串 V，如果用 abc 把字符串 V 切成 2 半，X 和 Y，组成 X + abc + Y 的字符串，X + abc + Y 的这个字符串依旧是有效的。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)这些都不是有效的字符串。
任意给一个字符串 S ，要求判断它是否有效，如果有效则输出 true。
解题思路 #
这一题可以类似括号匹配问题，因为 “abc” 这样的组合就代表是有效的，类似于括号匹配，遇到 “a” 就入栈，当遇到 “b” 字符的时候判断栈顶是不是 “a”，当遇到 “c” 字符的时候需要判断栈顶是不是 “a” 和 “b”。最后如果栈都清空了，就输出 true。
代码 #
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
}