14, and S is the number of two:
Description Title
Enter a sorted array and incrementing a number S, find the two numbers in the array, and so that they are exactly S, and if a plurality of digits equal to S, the product of the output of the minimum number of two.
Output Description:
Corresponding to each test case, the output of two numbers, the first small output.
Note: Write your own add debugging out, but a lot of complexity, see rankings, a good compact
Method 1:
# -*- coding:utf-8 -*-
class Solution:
def FindNumbersWithSum(self, array, tsum):
# write code here
# 因为是递增数组，所以最靠两边的一对就是积最小的
l = []
for x in array:
l.append(tsum-x)
for y in l:
if y in array:
return [tsum-y,y]
return []
Method 2: own, too concise
# -*- coding:utf-8 -*-
class Solution:
def FindNumbersWithSum(self, array, tsum):
# write code here
ans = []
i = 0
j = len(array)-1
while i < len(array)-1 and j > 0:
res = []
if array[i] + array[j] == tsum:
res.append(array[i])
res.append(array[j])
ans.append(res)
i += 1
j -= 1
if array[i] + array[j] > tsum:
j -= 1
if array[i] + array[j] < tsum:
i += 1
if len(ans) == 0:
return []
else:
return ans[0]
15, seeking 1 + 2 + 3 + ... + n, and
Description Title
seek 1 + 2 + 3 + ... + n, requires multiplication and division can not be used, for, while, if, else , switch, case and keywords such as conditional statement (A B:? C).
# -*- coding:utf-8 -*-
class Solution:
def Sum_Solution(self, n):
# write code here
return n and (n + self.Sum_Solution(n-1))
# -*- coding:utf-8 -*-
class Solution:
def __init__(self):
self.sum = 0
def Sum_Solution(self, n):
# write code here
def qiusum(n):
self.sum += n
n -= 1
return n>0 and self.Sum_Solution(n)
qiusum(n)
return self.sum