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)
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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).
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Space complexity : O(logn)O(logn)O(logn). Sorting takes O(logn) space.
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; } }