leetcode611. Valid Triangle Number

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本文介绍了一种算法，用于计算给定整数数组中能构成三角形的三元组数量。通过排序和双指针技巧，高效地解决了问题，并提供了两种实现方法。

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].

解法

public class Solution {
    public int triangleNumber(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        Arrays.sort(nums);
        int i = 2, j = 0, k = 0, count = 0;
        while (i < nums.length) {
            j = 0;
            k = i - 1;
            while (j < k) {
                if (nums[j] + nums[k] <= nums[i]) {
                    j++;
                } else if (nums[j] + nums[k] > nums[i] && Math.abs(nums[j] - nums[k]) < nums[i]) {
                    count += k - j;
                    k--;
                }
            }
            i++;
        }

        return count;
    }
}

解法二

排序+双指针法，求两个数之和大于target的个数

public class Solution {
    /*
     * @param nums: A list of integers
     * @return: An integer
     * @time: O(nlogn + n^2)
     * @space: O(1)
     */
    public int triangleNumber(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        Arrays.sort(nums);
        int ans = 0;

        for (int i = 0; i < nums.length; i++) {
            int left = 0, right = i - 1;
            while (left < right) {
                if (nums[left] + nums[right] > nums[i]) {
                    ans += right - left;
                    right--;
                } else {
                    left++;
                }
            }
        }

        return ans;
    }
}