Largest palindrome product——查找最大回文数

Description:

Find the largest palindrome made from the product of two n-digit numbers.

Since the result could be very large, you should return the largest palindrome mod 1337.

Example:

Input: 2

Output: 987

Explanation: 99 x 91 = 9009, 9009 % 1337 = 987

Note:

The range of n is [1,8]. **

一开始想到的是从最大乘数开始（递减），乘以另一个数（嵌套递减），判断乘积是否为回文数，此法可行但时间复杂度高。

参考别人的solution后发现一种更快的思路：
1.两数的乘积有最大值
2.从最大乘积开始递减，判断是否有乘数因子（只有一层循环）
3.自己构建回文数（两n位数乘积得到的回文数一定是偶数位，因此可利用左半部分构建）
4.注意特殊情况（n=1时，直接返回9）

代码块

public class Solution {
    public int largestPalindrome(int n) {
        if(1==n) return 9;

        int upBound = (int)Math.pow(10,n) - 1;
        int lowBound = (int)(upBound/10);

        long maxValue = (long)upBound*(long)upBound;
        // left part of multiply(palindrome)
        int leftHalf = (int)(maxValue / (long)Math.pow(10,n));

        long palindrome = 0;
        boolean palindromeFound = false;

        while(false == palindromeFound){
            palindrome = createPalindrome(leftHalf);
            // judge if there are two factors
            // Espacially note: long is different from int
            for(long i=upBound; i>lowBound; i--){
                // find the two factors
                if(0 == palindrome%i){
                    palindromeFound = true;
                    break;
                }
                if(i*ibreak;
                }
            }
            leftHalf--; // next palindrome
        }
        return (int)(palindrome % 1337);
    }//largestPalindrome
    private long createPalindrome(long left){
        String leftStr = Long.toString(left);
        String rightStr = new StringBuffer(leftStr).reverse().toString();
        return Long.parseLong(leftStr+rightStr);
    }//createPalindrome
}

注意： java中的long和int型变量进行运算时最好统一类型再运算，被坑惨了…