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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.  (Xor 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: trueExplanation: 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: trueExplanation: "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.

Approach #1: Stack. [Java]

class Solution {    public boolean isValid(String S) {        Stack
    
      stack = new Stack<>();        for (int i = 0; i < S.length(); ++i) {            if (S.charAt(i) == 'c') {                if (stack.empty() || stack.pop() != 'b') return false;                if (stack.empty() || stack.pop() != 'a') return false;            } else stack.push(S.charAt(i));        }        return stack.empty();    }}
    

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Analysis:

Keep a stack, whenever meet character of 'c', pop 'b' and 'a' at the end of the stack. Otherwise, return false;

Reference: