2017上海金馬五校程式設計競賽 E:Find Palindrome
阿新 • • 發佈:2019-02-05
Time Limit: 1 s
Description
Given a string S, which consists of lowercase characters, you need to find the longest palindromic sub-string. A sub-string of a string S is another string S' that occurs "in" S. For example, "abst" is a sub-string of "abaabsta". A palindrome is a sequence of characters which reads the same backward as forward.Input
There are several test cases. Each test case consists of one line with a single string S (1 ≤ |S | ≤ 50).Output
For each test case, output the length of the longest palindromic sub-string.
Sample Input
sasadasa bxabx zhuyuan
Sample Output
7 1 3
題目意思:
給你一串字元,讓你找出其中最長的迴文子串長度。
解題思路:
首先先看字串,對於第 i 個字元str[i],如果迴文子串有奇數個字元,那麼以 i 為中心向左向右擴充套件,直到發現對稱位置字元不相等,如果掃過X個字元,那麼迴文子串長度為2*X+1.
如果迴文子串有偶數個字元,對於第 i 個字元str[i]在最接近對稱軸的左邊。對於這種情況我們要從第 i 個位置開始掃描,並與第 i+1 個位置字元進行比較,在向左向右擴充套件,直到對稱位置字元不相同為止。如果掃過了X個字元,那麼迴文子串長度為2*X。
#include<iostream> #include<stdio.h> #include<algorithm> #include<string.h> #define INF 999999 using namespace std; int main() { int maxlen,i,j,len; ///最長對稱子串長度。 char str[1003]; while(gets(str)) { maxlen = 0; len = strlen(str); for(i = 0; i < len; i++) { ///考慮是i是奇數串的中心,以i為中心,同時往左往右擴充套件 for(j = 0; i-j>=0&&i+j<=len; j++) { if(str[i-j]!=str[i+j]) break; if(2*j+1>maxlen) maxlen = 2*j+1; } ///i是偶數串的中心 for(j = 0; i-j>=0&&i+j+1<len;j++) { if(str[i-j]!=str[i+j+1]) break; if(2*j+2>maxlen) maxlen = 2*j+2; } } printf("%d\n",maxlen); } return 0; }