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 } }
611. Valid Triangle Number (solution 1)
原文:https://www.cnblogs.com/johnnyzhao/p/14060665.html