leetcode 611. 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].

排序之后，从数字末尾开始往前遍历，将left指向首数字，将right之前遍历到的数字的前面一个数字，然后如果left小于right就进行循环，循环里面判断如果left指向的数加上right指向的数大于当前的数字的话，那么right到left之间的数字都可以组成三角形，这是为啥呢，相当于此时确定了i和right的位置，可以将left向右移到right的位置，中间经过的数都大于left指向的数，所以都能组成三角形，就说这思路叼不叼！加完之后，right自减一，即向左移动一位。如果left和right指向的数字之和不大于nums[i]，那么left自增1，即向右移动一位，参见代码如下：

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>#include <regex>using namespace std;class Solution{public:    int triangleNumber(vector<int>& nums)     {        if (nums.size() < 3)            return 0;        int count = 0, n = nums.size();        sort(nums.begin(), nums.end());        for (int i = n - 1; i >= 2; i--)        {            int left = 0, right = i - 1;            while (left < right)            {                if (nums[left] + nums[right] > nums[i])                {                    count = count + (right-left);                    right--;                }                else                    left++;            }        }        return count;    }};