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:

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].

思路：首先滿足三角形的條件就是任意兩邊之和大於第三邊；

array.sort之後，a < b <  c, 所以，a + c > b, b + c > a 肯定成立，那麼就要判斷 a + b > c的個數有多少個；

那麼我們就需要固定c，然後求a, b 的組合有多少個，那麼就是雙指針的算法；

如果a + b < c,  由於 a < b, 所以只能 a ++;

如果a + b > c, 那麼count += b - a; 如果a往右邊移動也是> c, 所以b 應該往左移動;

class Solution {
    public int triangleNumber(int[] nums) {
        if(nums == null || nums.length < 3) {
            return 0;
        }
        Arrays.sort(nums);
        int count = 0;
        for(int i = 2; i < nums.length; i++) {
            int left = 0; int right = i - 1;
            while(left < right) {
                if(nums[left] + nums[right] > nums[i]) {
                    count += right - left;
                    right--;
                } else {
                    left++;
                }
            }
        }
        return count;
    }
}