原题链接在这里：https://leetcode.com/problems/valid-triangle-number/

题目：

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:

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

题解：

也是Two Pointers, 构成三角要求三条边成a + b > c. 设nums[i] 为c, 前面的数中选出nums[l]为a, nums[r]为b.

若是nums[l] + nums[r] > c则符合要求，若继续向右移动l, 有r-l种组合都包括num[r]. 所以res += r-l. r左移一位, 之后可能还有符合条件的组合.

若是nums[l] + nums[r] <= c, 则向右移动l.

Time Complexity: O(n^2). sort 用时O(nlogn), 对于每一个c, Two Points用时O(n). n = nums.length.

Space: O(1).

AC Java:

 class Solution {

     public int triangleNumber(int[] nums) {

         if(nums == null || nums.length == 0){

             return 0;

         }

         int res = 0;

         Arrays.sort(nums);

         for(int i = 2; i<nums.length; i++){

             int l = 0;

             int r = i-1;

             while(l<r){

                 if(nums[l] + nums[r] > nums[i]){

                     res += r-l;

                     r--;

                 }else{

                     l++;

                 }

             }

         }

         return res;

     }

 }

类似3Sum Smaller.