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].
思路:
首先将数组排序。依次取出前两个边的长度。第三条边长度小于前两个边的和。
查找第三条边用二分查找。
int triangleNumber(vector<int>& a) { int i,j,k,n,left,right,mid,ans,sum; ans=0; sort(a.begin(),a.end()); n=a.size(); for (i=0;i<n;i++) for (j=i+1;j<n;j++) { sum=a[i]+a[j]; left=j;right=n; while (right-left>1) { mid=(left+right)/2; if (a[mid]>=sum) right=mid; else left=mid; } ans+=left-j; } return ans; }
参考:
https://leetcode.com/lympanda/
时间: 2024-10-29 10:46:20