3Sum Leetcode Python

标签：leetcode   python   array   

    (-1, -1, 2)

第一种做法是DFS 其解是NP hard是通解所有sub array的问题，所以过不了test但是自己机子是可以跑的。先贴在这里，

class Solution:
    # @return a list of lists of length 3, [[val1,val2,val3]]
    def dfs(self,num,valuelist,start,solution):
        if  len(valuelist)==3 and valuelist not in solution:
            solution.append(valuelist)
        for index in range(start,len(num)):
            valuelist=valuelist+[num[index]]
            self.dfs(num,valuelist,index+1,solution)
            valuelist=valuelist[:len(valuelist)-1]
    def threeSum(self, num):
        solution=[]
        num.sort()
        self.dfs(num,[],0,solution)
        return solution
        
        

当num[i]+num[left]+num[right]==0时候就得到想要的解。否则当和小于0时left=left+1 和大于0则right=right-1

This method‘s time complexity is O(n^2), we need to sort the array first, then we use three pointers, the first pointer is the index from 0 to len(num-1)

the second pointer is left=i+1 the third pointer is right=len(num)-1, while left<right we check the num[i]+num[left]+num[right] if the value is 0 we add it to the solution.

if sum is less than 0 we increase left otherwise decrease right.

class Solution:
    # @return a list of lists of length 3, [[val1,val2,val3]]
    def threeSum(self, num):
        solution=[]
        num.sort()
        
        for i in range(len(num)-1):
            if i>0 and num[i]==num[i-1]:
                continue
            left=i+1
            right=len(num)-1
            while left<right:
                val=num[i]+num[left]+num[right]
                if val==0 and [num[i],num[left],num[right]] not in solution:
                    solution.append([num[i],num[left],num[right]])
                    left=left+1
                    right=right-1
                elif val<0:
                    left=left+1
                else:
                    right=right-1
        return solution
        

3Sum Leetcode Python

标签：leetcode   python   array   

原文地址：http://blog.csdn.net/hyperbolechi/article/details/42803209