【leetcode】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].

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

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