5. Longest Palindromic Substring\224. Basic Calculator\string\372. Super Pow

5. Longest Palindromic Substring

题目描述：

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: "babad"

Output: "bab

Note: “aba” is also a valid answer.
Example:

Input: "cbbd"

Output: "bb"

找到最长的回文子串。

考虑子串是奇偶的情况，根据代码优化原则，一般来说减少分支预测，可以加快速度，所以这里我把一个奇偶循环拆成了两个。最后速度超过66%的c++。

class Solution {
public:
    string longestPalindrome(string s) {
        int slen = s.length();
        string res;
        if(!slen) return res;

        int max_cnt = 1;
        res = s[0];

        for(int i = 0; i < slen; i++) {
            int ind1 = i, ind2 = i;
            int cnt = -1;
            while(1) { 
                if(s[ind1] == s[ind2]) cnt += 2;
                else break;
                ind1--; ind2++;
                if(ind1 < 0 || ind2 >= slen) break;
            }
            if(cnt > max_cnt){
                ind1 = ind1 + 1;
                max_cnt = cnt; res = s.substr(ind1, max_cnt);
            }
        }
        for(int i = 0; i < slen-1; i++) {
            int ind1 = i, ind2 = i+1;
            int cnt = 0;
            while(1) {
                if(s[ind1] == s[ind2]) cnt += 2;
                else break;
                ind1--; ind2++;
                if(ind1 < 0 || ind2 >= slen) break;
            }
            if(cnt > max_cnt) {
                ind1 = ind1 + 1;
                max_cnt = cnt;  res = s.substr(ind1, max_cnt);
            }    
        }
        return res;
    }
};

获取子串的方法： string.substr(stt_point, num)

224. Basic Calculator

题目描述：

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .

You may assume that the given expression is always valid.

Some examples:

"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23

Note: Do not use the eval built-in library function.

代码实现：

372. Super Pow

在这里使用了一个数学知识：

ab % k = (a%k)(b%k)%k

$$a^{1234567} \mod k = (a^{1234560} \mod k) * (a^7 \mod k) \mod k = (a^{123456} \mod k)^{10} \mod k * (a^7 \mod k) \mod k$$

举一个例子：

f(a,1234567) = f(a, 1234560) * f(a, 7) % k = f(f(a, 123456),10) * f(a,7)%k;

下面的代码是从leetcode的discussion看到的，写得非常清楚，而且写得很巧妙，令人赏心悦目。

class Solution {
private:    
    const int base = 1337;
    int powmod(int a, int k) //a^k mod 1337 where 0 <= k <= 10
    {
        a %= base;
        int result = 1;
        for (int i = 0; i < k; ++i)
            result = (result * a) % base;
        return result;
    }
public:
    int superPow(int a, vector<int>& b) {
        if (b.empty()) return 1;
        int last_digit = b.back();
        b.pop_back();
        return powmod(superPow(a, b), 10) * powmod(a, last_digit) % base;
    }
};

非递归的方式可以写成如下：

class Solution {
const int c = 1337;
public:
    int superPow(int a, vector<int>& b) {
        int r = 1;
        a %= c;
        for(int i=0;i<b.size();i++){
            r = (powmod(r,10)*powmod(a,b[i]))%c;
        }
        return r;
    }
private:
    int powmod(int r,int k) {
        int x = 1;
        for(int i=1;i<=k;i++){
            x = (x*r) % c;
        }
        return x;
    }
};

参考链接：

https://discuss.leetcode.com/topic/50489/c-clean-and-short-solution