zoukankan      html  css  js  c++  java
  • 1003. Check If Word Is Valid After Substitutions

    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: 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.

    Approach #1: Stack. [Java]

    class Solution {
        public boolean isValid(String S) {
            Stack<Character> 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();
        }
    }
    

      

    Analysis:

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

    Reference:

    https://leetcode.com/problems/check-if-word-is-valid-after-substitutions/discuss/247626/JavaPythonC%2B%2B-Stack-Solution-O(N)

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    junit单元测试
    方法引用
    方法引用表达式(1)
    Stream流的常用方法
    Stream流
    综合案例:文件上传
    tcp通信协议
    python 生成器与迭代器
    Python 序列化与反序列化
    python 文件操作
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10799417.html
Copyright © 2011-2022 走看看