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Commit 6ef00cc

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Add C++ solutions for leetcode 949.
1 parent fd199fe commit 6ef00cc

4 files changed

+225
-0
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#include<vector>
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#include<algorithm>
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#include<iostream>
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using namespace std;
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class Solution {
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public:
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/**
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* 4个数字全排列总共有24种可能, 判断每一种可能是否能组成合法时间值, 如果能, 再和当前保存的最大值进行比较;
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* 最大值是一个int值, 用来表示分钟数;
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*/
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string largestTimeFromDigits(vector<int>& arr) {
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int largest = -1;
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// 依次暴力枚举出4个数
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for(int i = 0; i < 4; i++) {
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for(int j = 0; j < 4; j++) {
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for(int k = 0; k < 4; k++) {
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if(i == j || i == k || j == k) continue;
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// 0,1,2,3 总和为6, 故剩下的index为6-three sum of index
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int l = 6 - i - j - k;
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int res = composeTime(i, j, k, l, arr);
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largest = max(largest, res);
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}
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}
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}
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if(largest == -1) return "";
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string str1(5, '0');
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sprintf(&str1[0], "%02d:%02d", largest / 60, largest % 60);
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return str1;
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}
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/* 判断输入的四个数字按照输入顺序组成的时间是否合法, 如果合法, 返回分钟数 */
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int composeTime(int a, int b, int c, int d, vector<int>& nums)
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{
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int hours = nums[a] * 10 + nums[b];
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int min = nums[c] * 10 + nums[d];
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if(hours < 24 && min < 60)
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return hours * 60 + min;
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return -1;
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}
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};
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// Test
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int main()
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{
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Solution sol;
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vector<int> nums = {2, 0, 6, 6};
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auto res = sol.largestTimeFromDigits(nums);
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cout << res << endl;
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return 0;
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}
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#include<vector>
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#include<algorithm>
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#include<iostream>
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using namespace std;
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class Solution {
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public:
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string largestTimeFromDigits(vector<int>& arr) {
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sort(arr.begin(), arr.end());
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if (arr[0]*10 + arr[1] > 23) return "";
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vector<vector<int>> hhs; // hour, i, j
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for (int i = 0; i < 4; i++)
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{
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for (int j = 0; j < 4; j++)
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{
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if (i != j && (arr[i]*10 + arr[j] < 24))
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hhs.push_back({arr[i]*10 + arr[j], i, j});
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}
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}
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sort(hhs.begin(), hhs.end(), greater<>());
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vector<int> validHhs;
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vector<int> last2num;
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bool ok = getValidLast2num(hhs, arr, last2num, validHhs);
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if (!ok) return "";
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int hour = validHhs.at(0);
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int digitB = hour % 10;
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int digitA = hour / 10;
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if (last2num[0] >= 6 && last2num[1] >= 6) return "";
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string res = to_string(digitA) + to_string(digitB) + ":";
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string minu;
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if (last2num[0] < 6 && last2num[1] < 6)
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{
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sort(last2num.begin(), last2num.end(), greater<>());
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minu.push_back(last2num[0] + '0');
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minu.push_back(last2num[1] + '0');
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res += minu;
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}
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else if (last2num[0] >= 6 || last2num[1] >= 6)
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{
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sort(last2num.begin(), last2num.end());
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minu.push_back(last2num[0] + '0');
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minu.push_back(last2num[1] + '0');
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res += minu;
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}
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return res;
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}
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bool getValidLast2num(vector<vector<int>> hhs, vector<int> arr, vector<int>& last2num, vector<int>& validHhs)
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{
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bool canGetValidLast2num = false;
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for (int k = 0; k < hhs.size(); k++)
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{
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vector<int> hh = hhs[k];
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last2num.clear(); // 这个地方很关键
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for (int i = 0; i < 4; i++)
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{
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if (i != hh.at(1) && i != hh.at(2))
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last2num.push_back(arr[i]);
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}
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if ((last2num[0]*10 + last2num[1] <= 60) || (last2num[1]*10 + last2num[0] <= 60))
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{
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canGetValidLast2num = true;
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validHhs = hh;
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break;
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}
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}
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return canGetValidLast2num;
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}
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};
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// Test
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int main()
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{
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Solution sol;
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vector<int> nums = {2, 0, 6, 6};
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auto res = sol.largestTimeFromDigits(nums);
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cout << res << endl;
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return 0;
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}
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#include<vector>
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#include<algorithm>
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#include<iostream>
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using namespace std;
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class Solution {
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public:
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string largestTimeFromDigits(vector<int>& A) {
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string res = "";
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sort(A.begin(), A.end());
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do
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{
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if ((A[0] == 2 && A[1] <= 3 || A[0] < 2) && A[2] <= 5)
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{
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string curTime = to_string(A[0]) + to_string(A[1]) + ":"
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+ to_string(A[2])+to_string(A[3]);
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if (curTime > res) res = curTime;
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}
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} while (next_permutation(A.begin(), A.end()));
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return res;
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}
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};
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// Test
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int main()
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{
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Solution sol;
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vector<int> nums = {2, 0, 6, 6};
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auto res = sol.largestTimeFromDigits(nums);
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cout << res << endl;
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return 0;
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}
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#include<vector>
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#include<map>
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#include<regex>
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#include<algorithm>
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#include<iostream>
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using namespace std;
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class Solution {
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public:
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string res = "";
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string largestTimeFromDigits(vector<int>& A)
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{
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regex pattern("([0-1][0-9]:[0-5][0-9])|([2][0-3]:[0-5][0-9])");
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map<int, int> dict; // 存储每个数字digit出现的次数
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for (int i = 0; i < A.size(); i++)
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dict[A[i]]++;
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string str = "";
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backTrack(dict, str, pattern);
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return res;
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}
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void backTrack(map<int, int> dict, string str, regex& pattern)
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{
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if (str.length() == 2)
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str += ":";
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if (str.length() == 5)
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{
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if (regex_match(str, pattern))
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res = max(res, str); //valid time
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return;
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}
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for (auto it = dict.begin(); it != dict.end(); it++)
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{
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if (it->second != 0)
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{
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int val = it->first;
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dict[val]--;
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backTrack(dict, str + to_string(val), pattern);
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dict[val]++;
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}
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}
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}
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};
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// Test
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int main()
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{
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Solution sol;
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vector<int> nums = {2, 0, 6, 6};
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auto res = sol.largestTimeFromDigits(nums);
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cout << res << endl;
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return 0;
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}

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