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].
class Solution: def triangleNumber(self, nums): newlist = [ele for ele in nums if ele>0] #列表中的元素必须大于0 newlist.sort() length = len(newlist) count = 0 for k in reversed(range(0,length)): #从最大的数开始 i = 0 j = k-1 while i<j: if newlist[i]+newlist[j]>newlist[k]: #如果[i]+[j]>[k] count += j-i #说明i到j-1的元素与j组合都符合要求，共j-i种 j-=1 #说明我们可以适度减小[i]+[j]，去i~j-1内找符合要求的数 else: i+=1 #如果不符合要求，那么我们必须增大[i]+[j] return count