1.
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
分析:[x,y] ——>x+y=9,复杂度O(n**2);y=9-x...O(1)....loops....O(n),复杂度O(n)
代码:
class Solution:
def twoSum(self,nums,target):
dict1 = {nums[i]:i for i in range(len(nums))}
dict2 = {i:target-nums[i] for i in range(len(nums))}
twoSumShow = []
for i in range(len(nums)):
j = dict1.get(dict2.get(i))
if (j is not None) and (j != i):
twoSumShow = [i ,j]
break
return twoSumShow
2.
For example, given array S = {-1 0 1 2 -1 -4},
A solution set is:
(-1, 0, 1)
(-1, -1, 2)
分析1:(x,y,z)——>x+y+z=0,复杂度O(n**3);c = -(a+b)..set..O(1).....复杂度O(n**2)
代码:
def treeSum(self,nums):
if len(nums) < 3:
return []
nums.sort()
#sort方法,有利于判断是否重复
res = set()
for i,v in enumerate(nums[:-2]):
#v是第一层遍历,先枚举v
if i >= 1 and v == nums[i-1]:
continue
d = {}
for x in nums[i+1:]:
#再枚举x
if x not in d:
d[-v-x] = 1
#判断-v-x是否存在
else:
res.add((v,-v-x,x))
return map(list,res)
#map去重后返回出去
分析2:sort, find,快排:Array,sort,....O(NlogN);有序的,然后loops:a,b从左边开始,c从右边开始
代码:
def threeSum(self,nums):
res = []
nums.sort()
for i in xrange(len(nums)-2):
if i > 0 and nums[i] == nums[i-1]:
continue
l,r = i+1,len(nums)-1
while l < r:
s = nums[i] + nums[l] + nums[r]
if s < 0:
l +=1
elif s > 0:
r -=1
else:
res.append((nums[i],nums[l],nums[r]))
while l < r and nums[l] == nums[l+1]:
#判断是否重复
l +=1
while l < r and nums[r] == nums[r-1]:
#判断是否重复
r -= 1
l += 1; r -= 1
3.拓展(后续补充)
[1,2,3,4,-1,-2,-4,...n] ;return [a,b,c,d,...z] (a+b+c+d+..+z=0)
复杂度的减少,就是基本减少for循环。尽量减少遍历的次数。减少步骤。尽量向0和1的思维靠拢,毕竟电脑考虑的是0和1。