LeetCode Week5

EX.647

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

Input: "abc"Output: 3Explanation: Three palindromic strings: "a", "b", "c".

Example 2:

Input: "aaa"Output: 6Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".

Solution:

i为中间的，判断i的左右两个字母是否一样，如果一样，j++，直至两边不同，跳出内层循环。外层循环i++。

内层的两个循环分别对应奇数个字母的字符串和偶数个字母的字符串。

class Solution {public:    int countSubstrings(string s) {        int res = 0, n = s.length();        for(int i = 0; i < n; i++){            for(int j = 0; i-j >= 0 && i+j < n && s[i-j] == s[i+j]; j++)res++;            for(int j = 0; i-1-j >= 0 && i+j < n && s[i-1-j] == s[i+j]; j++)res++;        }        return res;    }};

EX.553

Given a list of positive integers, the adjacent integers will perform the float division. For example, [2,3,4] -> 2 / 3 / 4.

However, you can add any number of parenthesis at any position to change the priority of operations. You should find out how to add parenthesis to get the maximum result, and return the corresponding expression in string format. Your expression should NOT contain redundant parenthesis.

Example:

Input: [1000,100,10,2]Output: "1000/(100/10/2)"Explanation:1000/(100/10/2) = 1000/((100/10)/2) = 200However, the bold parenthesis in "1000/((100/10)/2)" are >redundant, since they don't influence the operation priority. So you >should return "1000/(100/10/2)". Other cases:1000/(100/10)/2 = 501000/(100/(10/2)) = 501000/100/10/2 = 0.51000/100/(10/2) = 2

Solution:
只要知道，在连除的式子中加括号时，把括号括起第二个数和最后一个数时，得到的结果最大即可，例如样例中的1000/(100/10/2)

class Solution {public:    string optimalDivision(vector<int>& nums) {        if (nums.size() == 0) return "";        string res = to_string(nums[0]);        if (nums.size() > 2) {            res += "/(";            for (int i = 1; i < nums.size()-1; i++) {                res += to_string(nums[i]);                res += "/";            }            res += to_string(nums[nums.size()-1]);            res += ")";        } else if (nums.size() == 2) {            res += "/";            res += to_string(nums[nums.size()-1]);        }        return res;    }};