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].
解析:
首先如何判断三个边长能否组成一个三角形,这个初中数学就学过——两个小边的和大于最大边的边长。
def checkValid(self,a,b,c): return (a+b) > c
因此,很容易得到解法,对nums数组排序后,做全排列,依次检查每种组合是否符合要求。
def triangleNumber2(self, nums): """ :type nums: List[int] :rtype: int """ count = 0 nums.sort() for i in range(0,len(nums)-2): for j in range(i+1,len(nums)-1): for k in range(j+1,len(nums)): if True == self.checkValid(nums[i], nums[j], nums[k]): count += 1 #print nums[i], nums[j], nums[k] else: break return count
但是非常遗憾,这种解法的复杂度是 O(n*3),超时了。根据判定中给出的超时输入发现,nums里面很多边长的数值是相同的。
Last executed input: [33,97,81,65,30,58,83,83,66,85,78,40,51,61,86,94,42,17,35,18,45,27,56,78,36,97,6,51,76,26,68,68,61,87,13,98,93,80,24,34,19,90,85,89,83,15,41,52,25,16,61,51,19,6,40,79,25,88,65,85,0,42,78,27,30,68,47,67,40,26,15,72,20,45,88,82,12,82,95,1,46,56,83,20,65,39,13,92,36,99,74,50,46,8,27,45,36,55,50,0]
因此可以考虑优化算法,首先统计三个边长都不一样的情况,然后再考虑等腰三角形的情况,最后是等边三角形。
def triangleNumber(self, nums): """ :type nums: List[int] :rtype: int """ #用一个list计数所有边长出线的次数,下标是边长值,对应的值是边长值出现的次数 count = 0 l = [] for i in range(0,1000+1): l.append(0) for i in nums: l[i] += 1 for i in range(1,len(l)-2): if l[i] == 0: continue for j in range(i+1,len(l)-1): if l[j] == 0: continue for k in range(j+1,len(l)): if l[k] == 0: continue if True == self.checkValid(i,j,k): count += (l[i]*l[j]*l[k]) else: break #dup for i in range(1,len(l)): if l[i] == 0 or l[i] == 1: continue else: #print ‘i:‘,i #等腰三角形,如果某个边长值的出线的次数为n(n>=2),表示可以组成等腰三角形。根据组合的原理,两个腰可以得到C(2,n) #然后根据 2*边长值与其他非自身边长值进行比较,得到等腰三角形的数量 for j in range(1,len(l)): if 2*i > j and l[j] > 0 and j != i: count += (l[i]*(l[i]-1))/2*l[j] #等边三角形,这个也是组合,如果某个边长值的出线的次数为n(n>=3),表示可以组成等边三角形。 #根据组合的原理,可以等到的种数为C(3,n) if l[i] >= 3: count += (l[i]*(l[i]-1)*(l[i]-2))/6 return count
这样优化之后,效率得到了很大的提升,但是还是不够。进一步检查发现,list中无效的元素太多,可以考虑过滤掉很多边长值出现次数为0的元素,如代码所示,用一个ul对l中的数据进行过滤。
def triangleNumber3(self, nums): """ :type nums: List[int] :rtype: int """ count = 0 l = [] for i in range(0,1000+1): l.append(0) for i in nums: l[i] += 1 #print l ul = [] for i in range(0,len(l)): if l[i] > 0: ul.append(i) #print ul for i in range(0,len(ul)-2): for j in range(i+1,len(ul)-1): for k in range(j+1,len(ul)): if True == self.checkValid(ul[i],ul[j],ul[k]): count += (l[ul[i]]*l[ul[j]]*l[ul[k]]) #print i,j,k else: break #dup #print count for i in range(0,len(ul)): if l[ul[i]] == 0 or l[ul[i]] == 1 or ul[i] == 0: continue else: #print ‘i:‘,i for j in range(0,len(ul)): if 2*ul[i] > ul[j] and l[ul[j]] > 0 and j != i and ul[j] != 0: #print (l[i]*(l[i]-1))/2 count += (l[ul[i]]*(l[ul[i]]-1))/2*l[ul[j]] #print count if l[ul[i]] >= 3: count += (l[ul[i]]*(l[ul[i]]-1)*(l[ul[i]]-2))/6 return count
这样再一优化之后,终于通过了。
时间: 2024-10-15 05:17:24