611. Valid Triangle Number (solution 1)

复制代码
package LeetCode_611

/**
 * 611. Valid Triangle Number
 * https://leetcode.com/problems/valid-triangle-number/
 * 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].
 * */
class Solution {
    /*
    Because Triangle is the sum of any two sides must be greater than third sides (任意两条边之和要大于第三边),
    we need to find 3 numbers,i<j<k and nums[i]+nums[j]>nums[k];
      Solution 1: sort array, bruce force, Time:O(n^3), Space:O(1);
    * Solution 2: sort array, two pointer, Time:O(n^2), Space:O(1);
    * */
    fun triangleNumber(nums: IntArray): Int {
        if (nums == null || nums.isEmpty() || nums.size < 2) {
            return 0
        }
        var count = 0
        nums.sort()
        //solution 1
        val n = nums.size
        for (i in 0 until n - 2) {
            for (j in i + 1 until n - 1) {
                val sum = nums[i] + nums[j]
                var k = j + 1
                while (k < n && sum > nums[k]) {
                    count++
                    k++
                }
            }
        }
        return count
    }
}
复制代码

 

posted @   johnny_zhao  阅读(57)  评论(0编辑  收藏  举报
编辑推荐:
· AI与.NET技术实操系列(二):开始使用ML.NET
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
阅读排行:
· DeepSeek 开源周回顾「GitHub 热点速览」
· 物流快递公司核心技术能力-地址解析分单基础技术分享
· .NET 10首个预览版发布:重大改进与新特性概览!
· AI与.NET技术实操系列(二):开始使用ML.NET
· .NET10 - 预览版1新功能体验(一)
点击右上角即可分享
微信分享提示