【剑指offer】和为S的连续正数序列（Python）

1. 暴力求解
2. 外部遍历到 tsum /2 就够了
3. 内部遍历到 count < tsum

# -*- coding:utf-8 -*-
class Solution:
    def FindContinuousSequence(self, tsum):
        # write code here
        if tsum < 3:
            return []
        result = []
        for i in range(1,tsum // 2 + 1):
            item = []
            count = 0
            while count < tsum:
                item.append(i)
                count += i
                i += 1
            if count == tsum:
                result.append(item)
        return result