[LeetCode]647. Palindromic Substrings

題目描述

思路

1. 簡單暴力遍歷
2. 減少比較次數，將外層迴圈遍歷的每個元素當作中間值，左右分別伸展，考慮奇數和偶數長度的字串情況

程式碼

#include <iostream>
#include <string>

using namespace std;

class Solution {
public:
    /*
    bool isPalind(string str, int start, int end) {
        while
 
 (start < end) {
            if (str[start] != str[end]) return false;
            ++start, --end;
        }
        return true;
    }

    int countSubstrings(string s) {
        int res = 0;
        for (int start = 0; start < s.size(); ++start)
            for (int end = start; end < s.size(); ++end
 
)
                if (isPalind(s, start, end)) ++res;
        return res;
    }
    */
    int countSubstrings(string s) {
        int res = 0;
        for (int mid = 0; mid < s.size(); ++mid) {
            for (int offset = 0; mid - offset >= 0 && mid + offset < s.size() && s[mid
 
 - offset] == s[mid + offset]; ++offset)
                ++res;
            for (int offset = 0; mid - offset - 1 >= 0 && mid + offset < s.size() && s[mid - offset - 1] == s[mid + offset]; ++offset)
                ++res;
        }
        return res;
    }
};

int main() {
    Solution s;

    cout << s.countSubstrings("aaa") << endl;

    system("pause");
    return 0;
}