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]

 #简单版本
 def Two_Sum(nums,target):
     i = 0
     j = 1
     while i < len (nums):
         while j < len(nums):
             if nums[i] + nums[j] == target:
                 print (i, j)
                 break
             j += 1
         i += 1

 nums = [2,7,11,15]
 target = int(input("输入target:"))

 #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
 #回溯法解决子集和数问题
 #两种形式：1.固定长度n-元组
 #2.可变元组
 def SumOfSum(s,k,r,x,M,w):
     x[k] = 1
     if s + w[k] == M:
         print(x)
     elif s + w[k] + w[k+1] <= M:#搜索左子树
         SumOfSum(s+w[k],k+1,r-w[k],x,M,w)
     elif s + r - w[k] >= M  and  s + w[k+1] <= M:#搜索右子树
         x[k] = 0
         SumOfSum(s,k+1,r-w[k],x,M,w)

 def SumOfSum1(x,M,w):
     r = 0
     for i in range(len(w)):
         r = r + w[i]
     print(r)#
     if r >= M and w[0] <= M:
         SumOfSum(0,0,r,x,M,w)

 n = 6
 x = [0] * 6
 w = [5, 10, 12, 13, 15, 18]
 M = 30
 SumOfSum1 (x, M, w)