标签：过滤 vector tin more public inpu start ace str
Given an array of 4 digits, return the largest 24 hour time that can be made.
The smallest 24 hour time is 00:00, and the largest is 23:59. Starting from 00:00, a time is larger if more time has elapsed since midnight.
Return the answer as a string of length 5. If no valid time can be made, return an empty string.
Example 1:
Input: [1,2,3,4] Output: "23:41"
Example 2:
Input: [5,5,5,5] Output: ""
Note:
A.length == 40 <= A[i] <= 9过滤掉不匹配的时间，然后更新最大时间即可。
#include <iostream> #include <algorithm> #include <vector> using namespace std; class Solution { public: string largestTimeFromDigits(vector<int>& a) { vector<int> b(4,-1); sort(a.begin(),a.end()); do { if(a[0]>2) continue; if(a[0]*10+a[1]>=24) continue; if(a[2]*10+a[3]>=60) continue; if(a[0]>b[0]||a[1]>b[1]||a[2]>b[2]||a[3]>b[3]) for(int i=0;i<4;++i) b[i]=a[i]; }while(next_permutation(a.begin(),a.end())); if(b[0]==-1) return ""; return to_string(b[0])+to_string(b[1])+":"+to_string(b[2])+to_string(b[3]); } }; int main() { Solution s; int a[4]={1,2,3,4}; vector<int> t(a,a+4); cout<<s.largestTimeFromDigits(t)<<endl; return 0; }
949. Largest Time for Given Digits
标签：过滤 vector tin more public inpu start ace str
原文地址：https://www.cnblogs.com/tianzeng/p/10054699.html
