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
A:
首先 all permuation . ------- 结果 LTE
class Solution { public: int largestPerimeter(vector<int>& A) { int res = 0; int n = A.size(); sort(A.begin(), A.end()); for(int i =0;i<n; i++){ for(int j = i+1;j<n;j++){ for(int k = j+1; k<n;k++){ if(A[i] + A[j] > A[k]){ res = max(res, A[i] + A[j] + A[k] ); } } } } return res; } };
再研读题目,发现我们应该从打的数开始比较。 而且只需要比较某个数之前2个位置就够了
class Solution { public: int largestPerimeter(vector<int>& A) { sort(A.begin(), A.end()); for(int i =A.size()-1;i>=2; i--){ if(A[i-2] + A[i-1] > A[i]) return A[i-2] + A[i-1] + A[i]; } return 0; } };
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