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:

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].

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

brute force当然可以，两边之和大于第三边

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

这种方法更简单，先sort，然后用两个指针，如果当前和>第三边，那么 right - left都大于第三边，然后就right--。如果相等或者小于了就left++

https://www.cnblogs.com/Moriarty-cx/p/9941005.html