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
}
}