Writing notes day4

Rituo   611 The number of valid triangles

 First of all, you need to know how to judge whether three numbers can form a triangle.

If there are three numbers a < b < c, if you want to form a triangle, you need to satisfy:

 a+b > c ; a + c > b ; b + c > a ; Any two numbers greater than the third can form a triangle.

In fact, it is not difficult to find that if the three numbers are sorted from largest to smallest, it only needs to satisfy a+b > c.

So the first step first sorts the given array.

Secondly, fix the largest number as a side, and then use the double pointer left to point to the first number, and right to point to the previous number of the fixed largest number. When nums[left] + nums[right] > the fixed maximum number, move the right pointer to the right; otherwise, move the pointer left to the left.

Up to the fixed maximum number is the end position of the third number of the array. (problem-solving ideas)

So how to count the number of effective triangles?

 

class Solution {
public:
    int triangleNumber(vector<int>& nums) {
        //先给数组排序
        sort(nums.begin(),nums.end());
        int c = nums.size()-1;
        int count = 0; // 记录有效三角形的个数
        while(c>=2)
        {
            int left = 0, right = c-1;
            while(left < right)
            {
                if(nums[left] +nums[right] > nums[c])
                {
                    count += (right - left);
                    --right;
                }else{
                    ++left;
                }
            }
            --c;
        }
        return count;
    }
};