#回溯练习
#给你两个整数 n和k，从1-n中选择k个数字的组合。比如n=4，那么从1,2,3,4中选取两个数字的组合，包括图上所述的四种[1, 2], [1, 3], [2, 3], [2, 4]。
def combine(n, k):
#global res_list
res_list = []
def backtrack(n, k, start, list_one):
if k == 0:
res_list.append(list_one[:])#注意，这里必须要[:]，为什么，不然都是记录引用，每次List_one改变。他的内容也是改变之后的内容？
#print(res_list)
print(list_one)
#return list_one
else:
for i in range(start, n+1):
list_one.append(i)
backtrack(n, k-1, i+1, list_one)
list_one.pop()
list_one = []
backtrack(n , k, 1, list_one)
print(res_list)
def combinationSum(candidates, target):
'''
找到数组中之和为target的所有组合
'''
def backtrack(candidates, remian, list_one, start):
#退出条件是剩余值为0和为负
if remian < 0:
return
if remian == 0:
result_list.append(list_one[:])
print(list_one)
#print(result_list)
else:
for i in range(start, len(candidates)):
remian = remian-candidates[i]
if remian < 0:
break
list_one.append(candidates[i])
backtrack(candidates, remian, list_one, i)
list_one.pop()
#千万注意也要把remian加回来
remian += candidates[i]
result_list = []
list_one = []
candidates.sort()
backtrack(candidates, target, list_one, 0)
return result_list
combinationSum([2, 3, 6, 7], 7)
