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Saturday, July 22, 2017

[Leetcode] Valid Triangle Number, Solution

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:
  1. The length of the given array won't exceed 1000.
  2. The integers in the given array are in the range of [0, 1000].

[Thoughts]

对于一个三角形,只要满足两边之和大于第三边即可。这题可采用双指针遍历。

首先把数组排序一遍,保证其有序。

其次,遍历数组,每一个数字都作为第三条边的选择,然后在前面的数字中通过双指针来决定第一条边和第二条边。对于任何一个可能的三角形,比如下例,3 + 7 > 9,那么如果第二条边(7)不变,所以第一条边(3)之后的数字都可以是解。




[Code]
1:    int triangleNumber(vector<int>& nums) {  
2:      if(nums.size() < 3) return 0;  
3:        
4:      sort(nums.begin(), nums.end());  
5:        
6:      int count = 0;  
7:      for(int third_edge =2; third_edge < nums.size(); third_edge++) {  
8:        int first_edge = 0, second_edge = third_edge-1;  
9:        while(first_edge < second_edge) {  
10:          if(nums[first_edge] + nums[second_edge] > nums[third_edge]) {  
11:            count+= second_edge - first_edge;  
12:            second_edge --;  
13:          }else {  
14:            first_edge ++;  
15:          }  
16:        }  
17:      }  
18:      return count;  
19:    }  


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