【LeetCode】611.Valid Triangle Number解题报告

【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;    }}

Date：2017年7月29日