【LeetCode】611.Valid Triangle Number解题报告
tags: Array
题目地址：https://leetcode.com/problems/valid-triangle-number/description/ 题目描述：
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.
Examples:
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].
Solutions:
解法一：暴力解法，三层循环，算出所有三个数的情况，显然超时。时间复杂度O(n³),空间复杂度O(1)。
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++; } } } I return count; } }解法二：排序—+二分查找，对输入数组nums排序，美剧长度最小的两条边，利用二分查找符合条件的最大边的下标。时间复杂度O(n²logn),空间复杂度O(logn)。
public class Solution { public int binarySearch(int[] nums, int start, int target) { int left = start, right = nums.length - 1; while (left <= right) { int mid = (left + right) / 2; if (nums[mid] >= target) right = mid - 1; else left = mid + 1; } return left; } public int triangleNumber(int[] nums) { Arrays.sort(nums); int size = nums.length; int ans = 0; for (int i = 0; i < size - 2; i++) { for (int j = i + 1; j < size - 1; j++) { int k = binarySearch(nums, j + 1, nums[i] + nums[j]); ans += k - j - 1; } } return ans; } }解法三：排序+双指针，对输入数组nums排序，枚举最小的边，利用双指针寻找符合条件的长度最大的两条边。时间复杂度O(n²),空间复杂度O(logn)。
public class Solution { public int triangleNumber(int[] nums) { Arrays.sort(nums); int size = nums.length; int ans = 0; for (int i = 0; i < size - 2; i++) { if (nums[i] == 0) continue; int k = i + 2; for (int j = i + 1; j < size - 1; j++) { while (k < size && nums[k] < nums[i] + nums[j]) k++; ans += k - j - 1; } } return ans; } }