976. Largest Perimeter Triangle--Easy

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

1.思考

• 定理：三角形任意两条边之和都大于第三边；
• 通过上面这条定理来确定某三条边是否能够组成三角形；
• 先对所有边进行长度的排序，从最长边开始，验证上面定理是否成立，从而得到周长。
  
  

2.实现
Runtime: 52ms(84.11%)
Memory: 10.5MB(98.15%)

class Solution {
public:    
    int largestPerimeter(vector<int>& A) {
        sort(A.begin(), A.end());
        int len = A.size();
        int res = 0;
        
        for(int i=len-3; i>=0 ; i--){
            if(A[i]+A[i+1]>A[i+2]){
                return A[i]+A[i+1]+A[i+2];
            }
        }
        return res;
    }
};

976. Largest Perimeter Triangle--Easy

原文：https://www.cnblogs.com/xuyy-isee/p/11303421.html