Java练习--三角形计算

最新推荐文章于 2024-08-31 10:30:00 发布

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本文介绍了一种高效算法，用于计算由非负整数组成的数组中能构成三角形的三元组数量。该算法首先对数组进行排序，接着采用双指针技术从前向后遍历数组，通过判断两短边之和是否大于最长边来确定能否构成三角形，复杂度为O(n²)。

Given an array consists of non-negative integers, your task is to count the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.

Example 1:

Input: [2,2,3,4]
Output: 3
Explanation:
Valid combinations are: 
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3

Note:

The length of the given array won't exceed 1000.
The integers in the given array are in the range of [0, 1000]

public class Solution {
    public int triangleNumber(int[] nums) {
        
    }
}

解决代码：

public static int triangleNumber(int[] A) {
    Arrays.sort(A);
    int count = 0, n = A.length;
    for (int i=n-1;i>=2;i--) {
        int l = 0, r = i-1;
        while (l < r) {
            if (A[l] + A[r] > A[i]) {
                count += r-l;
                r--;
            }
            else l++;
        }
    }
    return count;
}

先排序然后从大向前比较，小两边只和大于第三边则成立。复杂度：o(n2)