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].

    public int triangleNumber1(int[] nums) {
        Arrays.sort(nums);
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                for (int k = nums.length - 1; k >= 2 && k > j; k--) {
                    if (nums[i] + nums[j] > nums[k]) {
                        count++;
                    }
                }
            }
        }
        return count;
    }

    public int triangleNumber2(int[] nums) {
        Arrays.sort(nums);
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                for (int k = nums.length - 1; k >= 2 && k > j; k--) {
                    if (nums[i] + nums[j] > nums[k]) {
                        count = count + k - j;
                        break;
                    }
                }
            }
        }
        return count;
    }

    public int triangleNumber3(int[] nums) {
        if (nums.length < 3)
            return 0;
        Arrays.sort(nums);
        int res = 0;
        for (int i = nums.length - 1; i >= 0; i--) {
            int left = 0;
            int right = i - 1;
            while (right > left) {
                if (nums[left] + nums[right] > nums[i]) {
                    res += right - left;
                    right--;
                } else {
                    left++;
                }
            }
        }
        return res;
    }