Given an array A
of positive lengths, return the largest perimeter of a triangle with non-zero area, formed from 3 of these lengths.
If it is impossible to form any triangle of non-zero area, return 0
.
Example 1:
Input: [2,1,2] Output: 5
Example 2:
Input: [1,2,1] Output: 0
Example 3:
Input: [3,2,3,4] Output: 10
Example 4:
Input: [3,6,2,3] Output: 8
Note:
3 <= A.length <= 10000
1 <= A[i] <= 10^6
Solution: Greedy
Answer must be 3 consecutive numbers in the sorted array
if A[i] >= A[i+1] + A[i+2], then A[i] >= A[i+j] + A[i+k], 1 < j < k
if A[i] < A[i+1] + A[i+2], then A[i] + A[i+1] + A[i+2] is the answer
C++
1 2 3 4 5 6 7 8 9 10 |
class Solution { public: int largestPerimeter(vector<int>& A) { sort(rbegin(A), rend(A)); for (int i = 0; i < A.size() - 2; ++i) if (A[i] < A[i + 1] + A[i + 2]) return A[i] + A[i + 1] + A[i + 2]; return 0; } }; |