[Leetcode] Valid Triangle Number, Solution

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].

[Thoughts]

对于一个三角形，只要满足两边之和大于第三边即可。这题可采用双指针遍历。

首先把数组排序一遍，保证其有序。

其次，遍历数组，每一个数字都作为第三条边的选择，然后在前面的数字中通过双指针来决定第一条边和第二条边。对于任何一个可能的三角形，比如下例，3 + 7 > 9，那么如果第二条边（7）不变，所以第一条边（3）之后的数字都可以是解。

[Code]

1:    int triangleNumber(vector<int>& nums) {  
2:      if(nums.size() < 3) return 0;  
3:        
4:      sort(nums.begin(), nums.end());  
5:        
6:      int count = 0;  
7:      for(int third_edge =2; third_edge < nums.size(); third_edge++) {  
8:        int first_edge = 0, second_edge = third_edge-1;  
9:        while(first_edge < second_edge) {  
10:          if(nums[first_edge] + nums[second_edge] > nums[third_edge]) {  
11:            count+= second_edge - first_edge;  
12:            second_edge --;  
13:          }else {  
14:            first_edge ++;  
15:          }  
16:        }  
17:      }  
18:      return count;  
19:    }