标签：alt div sed pen cte turn public splay display
题目描述：
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".
解题思路：
本体主要是考察s中每个子字符串是否是回文字符串，有两种方法：
1.是确定子字符串的首尾，判断是否为回文字符串。
2.以每个字符为中心，向两端延伸判断是否相等，相等时回文字符串个数加一，不相等时终止内层循环。
相比较而言，第二种方法效率更高，可以更快终止遍历。
代码：
1 class Solution { 2 public: 3 int countSubstrings(string s) { 4 int ret = s.size(); 5 int j; 6 for (int i = 0; i < s.size(); i++){ 7 for (int j = i + 1; j < s.size(); ++j) { 8 if (s[i] != s[j]) 9 continue; 10 else 11 if (check(s, i, j)) 12 ret++; 13 } 14 } 15 return ret; 16 } 17 bool check(const string& s, int left, int right) { 18 int i = left; 19 int j = right; 20 for(;i <= j; i++, j--) { 21 if (s[i] != s[j]) 22 return false; 23 } 24 return true; 25 } 26 };
1 class Solution { 2 public: 3 int countSubstrings(string s) { 4 int ret = 0; 5 for (int i = 0; i < s.size(); i++) { 6 ret += check(s, i, i); 7 ret += check(s, i, i+1); 8 } 9 return ret; 10 } 11 int check(const string& s, int left, int right) { 12 int num = 0; 13 while (left>=0 && right<s.size() && s[left--]==s[right++]) 14 num++; 15 return num; 16 } 17 };
class Solution {public: int countSubstrings(string s) { int ret = 0; for (int i = 0; i < s.size(); i++) { ret += check(s, i, i); ret += check(s, i, i+1); } return ret; } int check(const string& s, int left, int right) { int num = 0; while (left>=0 && right<s.size() && s[left--]==s[right++]) num++; return num; }};
标签：alt div sed pen cte turn public splay display
原文地址：https://www.cnblogs.com/gsz-/p/9538491.html
