题解 | #最长回文子串#

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param A string字符串 
     * @return int整型
     */
    int condition1(int min_size,string A,int i){
        int id=1;
        int temp=1;
        while(min_size){
            if(A[i-id] == A[i+id]){
                temp+=2;
                id++;
                min_size--;
                // cout<<"hello"<<endl;
            }else{
                break;
            }
        }
        return temp;
    }
    int condition2(int min_size,string A,int i){
        int temp=0;
        int id=1;
        while(min_size){
            if(A[i-id+1] == A[i+id]){
                temp+=2;
                id++;
                min_size--;
            }else{
                break;
            }
        }
        return temp;
    }
    int getLongestPalindrome(string A) {
        // write code here
        int rel=1;
        for(int i=0;i<A.size()-1;i++){
            int min_size;
            //以i为中心向两边扩散
            if((i-0)<(A.size()-1-i)){
                min_size=i-0;
            }else{
                min_size=A.size()-1-i;
            }
            int temp1=condition1(min_size, A,i);
            //以i为边界比较两边的字符
            if((i-0+1)<(A.size()-1-i)){
                min_size=i-0+1;
            }else{
                min_size=A.size()-1-i;
            }
            int temp2=condition2(min_size, A,i);
            if(rel<temp1){
                rel=temp1;
            }
            if(rel<temp2){
                rel=temp2;
            }

        }
        return rel;

    }
};