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  • 647. Palindromic Substrings 互文的子字符串

    [抄题]:

    Given a string, your task is to count how many palindromic substrings in this string.

    The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

    Example 1:

    Input: "abc"
    Output: 3
    Explanation: Three palindromic strings: "a", "b", "c".
    

    Example 2:

    Input: "aaa"
    Output: 6
    Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".

     [暴力解法]:

    时间分析:

    空间分析:

     [优化后]:

    时间分析:

    空间分析:

    [奇葩输出条件]:

    [奇葩corner case]:

    [思维问题]:

    不知道和dp有什么关系:判断互文还是要用helper函数,dp只是写出由内而外的扩展方程

    [一句话思路]:

    由于自身就算互文串,所以自身扩展或相邻位扩展

    [输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

    [画图]:

    [一刷]:

    1. 递推表达式的范围正常写就行 不用-1 :for (int i = 0; i < s.length(); i++)

    [二刷]:

    [三刷]:

    [四刷]:

    [五刷]:

      [五分钟肉眼debug的结果]:

    [总结]:

    [复杂度]:Time complexity: O() Space complexity: O()

    [英文数据结构或算法,为什么不用别的数据结构或算法]:

    [算法思想:递归/分治/贪心]:

    [关键模板化代码]:

    递推表达式正常写就行,反正都由void类型的ispalindrome控制,一言不合就退出 

    public void isPalindromic(int left, int right, String s) {
            while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
                count++;
                left--;
                right++;
            }
        }

    [其他解法]:

    [Follow Up]:

    [LC给出的题目变变变]:

     [代码风格] :

    class Solution {
        int count = 0;
        
        public int countSubstrings(String s) {
            //cc
            if (s == null || s.length() == 0) return 0;
            
            //ini
            
            //for loop
            for (int i = 0; i < s.length(); i++) {
                isPalindromic(i, i, s);
                isPalindromic(i, i + 1, s);
            }
            
            return count;
        }
        
        public void isPalindromic(int left, int right, String s) {
            while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
                count++;
                left--;
                right++;
            }
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/immiao0319/p/9003299.html
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