原题链接在这里:https://leetcode.com/problems/valid-triangle-number/
题目:
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].
题解:
也是Two Pointers, 构成三角要求三条边成a + b > c. 设nums[i] 为c, 前面的数中选出nums[l]为a, nums[r]为b.
若是nums[l] + nums[r] > c则符合要求,若继续向右移动l, 有r-l种组合都包括num[r]. 所以res += r-l. r左移一位, 之后可能还有符合条件的组合.
若是nums[l] + nums[r] <= c, 则向右移动l.
Time Complexity: O(n^2). sort 用时O(nlogn), 对于每一个c, Two Points用时O(n). n = nums.length.
Space: O(1).
AC Java:
1 class Solution { 2 public int triangleNumber(int[] nums) { 3 if(nums == null || nums.length == 0){ 4 return 0; 5 } 6 7 int res = 0; 8 Arrays.sort(nums); 9 for(int i = 2; i<nums.length; i++){ 10 int l = 0; 11 int r = i-1; 12 while(l<r){ 13 if(nums[l] + nums[r] > nums[i]){ 14 res += r-l; 15 r--; 16 }else{ 17 l++; 18 } 19 } 20 } 21 return res; 22 } 23 }
类似3Sum Smaller.