LeetCode--Two_Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1]

 1 #简单版本
 2 def Two_Sum(nums,target):
 3     i = 0
 4     j = 1
 5     while i < len (nums):
 6         while j < len(nums):
 7             if nums[i] + nums[j] == target:
 8                 print (i, j)
 9                 break
10             j += 1
11         i += 1
12 
13 nums = [2,7,11,15]
14 target = int(input("输入target:"))

 1 #复杂版本--修改：已知n个不同整数wi的集合，求该集合的所有满足条件的子集，使得每个子集的正数之和等于另一个给定的正数M
 2 #回溯法解决子集和数问题
 3 #两种形式：1.固定长度n-元组
 4 #2.可变元组
 5 def SumOfSum(s,k,r,x,M,w):
 6     x[k] = 1
 7     if s + w[k] == M:
 8         print(x)
 9     elif s + w[k] + w[k+1] <= M:#搜索左子树
10         SumOfSum(s+w[k],k+1,r-w[k],x,M,w)
11     elif s + r - w[k] >= M  and  s + w[k+1] <= M:#搜索右子树
12         x[k] = 0
13         SumOfSum(s,k+1,r-w[k],x,M,w)
14 
15 def SumOfSum1(x,M,w):
16     r = 0
17     for i in range(len(w)):
18         r = r + w[i]
19     print(r)#73
20     if r >= M and w[0] <= M:
21         SumOfSum(0,0,r,x,M,w)
22 
23 n = 6
24 x = [0] * 6
25 w = [5, 10, 12, 13, 15, 18]
26 M = 30
27 SumOfSum1 (x, M, w)