Leetcode#1-2

#1 TowSum

class Solution:
    def twoSum(self, nums, target):
        
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        j=len(nums)-1
        ind = []
        indv = []
        r,l = 0,0
        minn = min(nums)
        for n,v in enumerate(nums):
            if v+minn<=target:
                ind.append(n)
                indv.append(v)
        for i in ind:
            if target-nums[i] in indv:
                indv.remove(nums[i])
                ind.remove(i)
                if target-nums[i] in indv:
                    l = i
                    x = indv.index(target-nums[i])
                    r = ind[x]
                    return [l,r]
        

 　　

　　执行用时: 740 ms, 在Two Sum的Python3提交中击败了54.94% 的用户

　　内存消耗: 7.4 MB, 在Two Sum的Python3提交中击败了79.99% 的用户

　　自己想的是这个方法，看了答案后，发现一个很神奇的函数　map(),传入的参数是map(function,list),返回值是一个新的list

#1 TowSum Answer

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        mirror = {}
        for idx, num in enumerate(nums):
            if num in mirror:
                return [mirror[num], idx]
            mirror[target - num] = idx

 

答案和我的思路是一样的,思维上更高阶,字典还不太熟练的运用.

 

#2 addTwoNumbers

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        b1 = l1
        b2 = l2
        b1l=[]
        b2l=[]
        r=[]
        while(b2):
            b2l.append(str(b2.val))
            b2 = b2.next
        while(b1):
            b1l.append(str(b1.val))
            b1 = b1.next
        n1 = eval(''.join(b1l[::-1]))
        n2 = eval(''.join(b2l[::-1]))
        ans = list(str(n1+n2))
        for i in range(len(ans)):
            r.append(eval(ans.pop(-1)))            
        return r
        
        
        
        
        

　　

执行用时: 232 ms, 在Add Two Numbers的Python3提交中击败了4.97% 的用户

内存消耗: 6.7 MB, 在Add Two Numbers的Python3提交中击败了91.98% 的用户

#2

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """

        add = 0
        l3 = ListNode(0)
        node = l3
        while l1 or l2:
            cur = ListNode(add)
            if l1:
                cur.val += l1.val
                l1 = l1.next
            if l2:
                cur.val += l2.val
                l2 = l2.next
            add = cur.val // 10
            cur.val = cur.val % 10
            node.next, node = cur, cur
        if add:
            node.next = ListNode(add)
        
        return l3.next