CodeSignal 刷題 —— almostIncreasingSequence

標籤：ISE   color   div   tin   pre   des   his   alt   ble   

Given a sequence of integers as an array, determine whether it is possible to obtain a strictly increasing sequence by removing no more than one element from the array.

Example

• For sequence = [1, 3, 2, 1], the output should be
  almostIncreasingSequence(sequence) = false.

  There is no one element in this array that can be removed in order to get a strictly increasing sequence.

• For sequence = [1, 3, 2], the output should be
  almostIncreasingSequence(sequence) = true.

  You can remove 3 from the array to get the strictly increasing sequence [1, 2]. Alternately, you can remove 2 to get the strictly increasing sequence [1, 3].

Input/Output

• [execution time limit] 0.5 seconds (cpp)

• [input] array.integer sequence

  Guaranteed constraints:
  2 ≤ sequence.length ≤ 105,
  -105 ≤ sequence[i] ≤ 105.

• [output] boolean

  • Return true if it is possible to remove one element from the array in order to get a strictly increasing sequence, otherwise return false.

 

C++解法：

 1 bool isSequence(vector<int> sequence) 2 { 3     set<int>iset; 4     for(auto e: sequence) 5         iset.insert(e); 6     if(iset.size() != sequence.size())     //如果sequence中有重複元素，返回false 7         return false; 8  9     vector<int>v = sequence;               //如果sequence與排序後的v相等，則說明sequence是遞增有序的10     sort(v.begin(), v.end());11     if(sequence == v)12         return true;13     else14         return false;15 16 }17 18 bool almostIncreasingSequence(std::vector<int> sequence) 19 {20     if (isSequence(sequence))21         return true;22     int i = 0;23     while (i < sequence.size())24     {25         vector<int> v = sequence;26         v.erase(v.begin() + i);27         if(isSequence(v))28             return true;29         ++i;30     }31     return false;32 }

 

CodeSignal 刷題 —— almostIncreasingSequence