Leetcode 之 Valid Triangle Number

611. Valid Triangle Number

1.Problem

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].

2.Solution

题目的大意是给定一个元素可能存在重复的一维数组，给出其中能组成合法三角形的组合数。

3.Code

//Solution1
//时间复杂度O(N^3),空间复杂度O(logN)(排序导致)
//先对nums数组进行排序，如从大小三条边为 a，b，c，只需判断 a + b > c 成立与否 =》三条边能否构成三角形
class Solution {
    public int triangleNumber(int[] nums) {
        Arrays.sort(nums);
        int l = nums.length;
        int count = 0;
        for ( int i = 0 ; i < l - 2 ; i++ ) {
            for ( int j = i + 1 ; j < l - 1 ; j++ ) {
                for ( int k = j + 1 ; k < l ; k++ ) {
                    if ( nums[i] + nums[j] > nums[k]) {
                        count++;
                    }
                }
            }
        }
        return count;
    }

}

//Solution2 二分查找
//时间复杂度O(N^2 * log(N))，空间复杂度O(log N)
class Solution {
    public int triangleNumber(int[] nums) {
        Arrays.sort(nums);
        int l = nums.length;
        int count = 0;
        for ( int i = 0 ; i < l - 2 ; i++ ) {
            int k = i + 2;
            for ( int j = i + 1 ; j < l - 1 && nums[i] != 0 ; j++ ) {
                k = binarySearch(k,l - 1,nums,nums[i] + nums[j]); 
                count += k - j - 1;   
            }
        }
        return count;
    }

    public int binarySearch( int start , int end , int[] nums , int target ) {
        while ( start <= end ) {
            int mid = ( end - start ) / 2 + start;
            if ( nums[mid] < target ) {
                start = mid + 1;
            } else {
                end = mid - 1;
            }
        }
        return start;
    }

}

//Solution3,时间复杂度O(N^2) ,空间复杂度O(lgN)

ime complexity : $O(n2)O(n^2)O(n2). Loop of kkk and jjj will be executed O(n2)O(n^2)O(n2) times in total, because, we do not reinitialize the value of kkk for a new value of jjj chosen(for the same iii). Thus the complexity will be O(n^2+n^2)=O(n^2).$
Space complexity :

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