LeetCode #976 Largest Perimeter Triangle 三角形的最大周长

976 Largest Perimeter Triangle 三角形的最大周长

Description:
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:

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

题目描述:
给定由一些正数(代表长度)组成的数组 A,返回由其中三个长度组成的、面积不为零的三角形的最大周长。

如果不能形成任何面积不为零的三角形,返回 0。

示例 :

示例 1:

输入:[2,1,2]
输出:5

示例 2:

输入:[1,2,1]
输出:0

示例 3:

输入:[3,2,3,4]
输出:10

示例 4:

输入:[3,6,2,3]
输出:8

提示:

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

思路:

先排序, 然后从最大值开始, 遍历到最大边小于另外两边之和, 这个三角形即为最大周长的三角形
时间复杂度O(nlgn), 空间复杂度O(1)

代码:
C++:

class Solution 
{
public:
    int largestPerimeter(vector& 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;
    }
};

Java:

class Solution {
    public int largestPerimeter(int[] A) {
        Arrays.sort(A);
        for (int i = A.length - 1; i >= 2; i--) if (A[i - 2] + A[i - 1] > A[i]) return A[i - 2] + A[i - 1] + A[i];
        return 0;
    }
}

Python:

class Solution:
    def largestPerimeter(self, A: List[int]) -> int:
        A.sort(reverse=True)
        for i in range(2, len(A)):
            if A[i] + A[i - 1] > A[i - 2]:
                return A[i - 2] + A[i - 1] + A[i]
        return 0

你可能感兴趣的:(LeetCode #976 Largest Perimeter Triangle 三角形的最大周长)