LeetCode Valid Triangle Number

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.