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