976. Largest Perimeter Triangle

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:

1. 3 <= A.length <= 10000
2. 1 <= A[i] <= 10^6

 

Approach #1: Math. [Java]

class Solution {
    public int largestPerimeter(int[] A) {
        
        Arrays.sort(A);
        
        int n = A.length;
       
        // Integer[] arr = new Integer[A.length];
        // for (int i = 0; i < A.length; ++i)
        //     arr[i] = A[i];
        
        // Arrays.sort(A, new Comparator<Integer>() {
        //     @Override
        //     public int compare(Integer a, Integer b) {
        //         return b - a;
        //     }
        // });
        
        // Arrays.sort(arr, (Integer a, Integer b) -> b - a);
        
        // Arrays.sort(arr, Collections.reverseOrder());

        for (int i = n - 1; i >= 0; --i) {
            for (int j = i - 1; j >= 0; --j) {
                for (int k = j - 1; k >= 0; --k) {
                    if (A[k] + A[j] > A[i]) {
                        return A[k] + A[j] + A[i];
                    }
                }
            }
        }

        return 0;
    }
}

　　

Analysis:

There three ways to realise reverse array, but the permise is that array's type must is object. so we have to change the type of A, If we do so, it will TEL. 

 

Approach #2: Math. [Java]

class Solution {
    public int largestPerimeter(int[] A) {
        int c = max(A);
        int b = max(A);
        int a = max(A);
        for(int i = 2; i < A.length; i++) {
            if(c < a + b)
                return a + b + c;
            else {
                c = b;
                b = a;
                a = max(A);
            }
        }
        return 0;
    }
    
    int max(int[] A) {
        int m = A[0];
        int index = 0;
        for(int i = 1; i < A.length; i++) {
            if(A[i] > m) {
                m = A[i];
                index = i;
            }
        }
        A[index] = -1;
        return m;
    }
}