LeetCode-1629. Key with the longest key duration

LeetCode Designed a new keyboard , Testing its availability . The tester will click a series of keys （ A total of n individual ）, One at a time .

Give you a length of n String keysPressed , among keysPressed[i] Represents the... In the test sequence i A pressed key .releaseTimes It's a list in ascending order , among releaseTimes[i] It means that the... Is released i Key time . String and array Subscripts are all from 0 Start . The first 0 Keys at time 0 When pressed , Next, each key just Pressed when the previous key is released .

The tester wants to find the key The longest duration Key . The first i The duration of the second key is releaseTimes[i] - releaseTimes[i - 1] , The first 0 The duration of the second key is releaseTimes[0] .

Be careful , Testing period , The same key can be pressed many times at different times , And the duration of each time may be different .

Please return to the key The longest duration Key , If there are multiple such keys , Then return to In alphabetical order, the largest The key of .

Example 1：

Input ：releaseTimes = [9,29,49,50], keysPressed = "cbcd"
Output ："c"
explain ： The key sequence and duration are as follows ：
Press down 'c' , The duration of the 9（ Time 0 Press down , Time 9 Release ）
Press down 'b' , The duration of the 29 - 9 = 20（ Time to release the last key 9 Press down , Time 29 Release ）
Press down 'c' , The duration of the 49 - 29 = 20（ Time to release the last key 29 Press down , Time 49 Release ）
Press down 'd' , The duration of the 50 - 49 = 1（ Time to release the last key 49 Press down , Time 50 Release ）
The key with the longest key duration is 'b' and 'c'（ When pressed the second time ）, The duration is 20
'c' Arrange in alphabetical order 'b' Big , So the answer is 'c'
Example 2：

Input ：releaseTimes = [12,23,36,46,62], keysPressed = "spuda"
Output ："a"
explain ： The key sequence and duration are as follows ：
Press down 's' , The duration of the 12
Press down 'p' , The duration of the 23 - 12 = 11
Press down 'u' , The duration of the 36 - 23 = 13
Press down 'd' , The duration of the 46 - 36 = 10
Press down 'a' , The duration of the 62 - 46 = 16
The key with the longest key duration is 'a' , The duration of the 16
 

Tips ：

releaseTimes.length == n
keysPressed.length == n
2 <= n <= 1000
1 <= releaseTimes[i] <= 109
releaseTimes[i] < releaseTimes[i+1]
keysPressed It's only made up of lowercase letters

#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>
using namespace std;
class Solution {
public:

	char slowestKey(vector<int>& releaseTimes, string keysPressed) {
		uint32_t maxVal = 0;
		vector<char> result;
		unordered_map<char, int> hash;
		unordered_map<char, int>::iterator it;

		hash[keysPressed[0]] = releaseTimes[0];
		maxVal = releaseTimes[0];
		for (int i = 1; i < keysPressed.size(); i++) {
			int val = releaseTimes[i] - releaseTimes[i - 1];
			if (hash.count(keysPressed[i]) == 0 || val > hash[keysPressed[i]]) {
				hash[keysPressed[i]] = val;
			}
			if (val > maxVal) {
				maxVal = val;
			}
		}
		it = hash.begin();
		while (it != hash.end()) {
			if (it->second == maxVal) {
				result.push_back(it->first);
			}
			it++;
		}
		sort(result.begin(), result.end(), cmp);
		return result[0];
	}

private:
	static bool cmp(char a, char b) {
		return (a >= b ? true : false);
	}
};

int main() {

	vector < int> v = { 23, 34, 43, 59, 62, 80, 83, 92, 97 };
	string str = "qgkzzihfc";
	unique_ptr<Solution> ptr(new Solution());
	cout << ptr->slowestKey(v, str) << endl;

	system("pause");
	return 0;
}