#第一种方法#
#时间换取空间:两个循环求解#
def search(nums,target):
for i in nums:
start = nums.index(i) + 1
for j in nums[start: ]:
if i + j == target:
result = [nums.index(i)]
next_index = nums[start:].index(j) + start
result.append(next_index)
return result
if __name__ == '__main__':
print(search([3,2,4],6))
#第二种方法#
#空间换取时间:引入字典快速求解#
def search(nums,target):
resut_list = list()
result_dict = {}
result_dict[nums[0]] = 0
for i in nums[1:]:
key = target - i
if key in result_dict.keys():
resut_list.append(result_dict[key])
resut_list.append(nums.index(i))
else:
result_dict[i] = nums.index(i)
return resut_list
if __name__ == '__main__':
print(search([3,2,4],6))
有兴趣的小伙伴也可以思考一下,如何返回数组里满足条件的所有值的组合。