摘要: 大伙都**挺猛的,怎么到你这就拉了垮了 阅读全文
posted @ 2020-08-15 15:52 珊珊来迟0 阅读(75) 评论(0) 推荐(0) 编辑
摘要: rsync: sudo rsync -avzr youliang@192.168.10.15:/home/youliang/code ./temp/ 阅读全文
posted @ 2020-10-21 11:05 珊珊来迟0 阅读(58) 评论(0) 推荐(0) 编辑
摘要: class Solution: def permute(self, nums: List[int]) -> List[List[int]]: res = [] def helper(nums, temp): if not nums: if temp not in res: res.append(te 阅读全文
posted @ 2020-10-09 21:35 珊珊来迟0 阅读(106) 评论(0) 推荐(0) 编辑
摘要: mosaic failed in semantic segmentaion 1. problem 为了PCL比赛,实现了mmseg的mosaic 经过mosaic ratio = 0的验证和dataloader的visualization,实现并未出现问题。 2. 猜想 segmentation注重 阅读全文
posted @ 2020-09-25 16:07 珊珊来迟0 阅读(180) 评论(0) 推荐(0) 编辑
摘要: 暴力 class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { int i, j; for (i = 0; i < nums.size(); ++i){ for(j = i + 1; j < nums.s 阅读全文
posted @ 2020-09-21 14:20 珊珊来迟0 阅读(88) 评论(0) 推荐(0) 编辑
摘要: torch scheduler class LRScheduler(object): r"""Learning Rate Scheduler Parameters mode : str Modes for learning rate scheduler. Currently it supports 阅读全文
posted @ 2020-09-18 10:56 珊珊来迟0 阅读(793) 评论(0) 推荐(0) 编辑
摘要: sigmoid focal loss class SigmoidFocalLoss(nn.Module): def __init__(self, ignore_label, gamma=2.0, alpha=0.25, reduction='mean'): super(SigmoidFocalLos 阅读全文
posted @ 2020-09-18 10:48 珊珊来迟0 阅读(407) 评论(0) 推荐(0) 编辑
摘要: 剑指 Offer 29. 顺时针打印矩阵 描述 略 思路 画好图,定义好(x1, y1, x2, y2)的物理意义 注意循环条件 注意终止位置 更新内外层 图片来自https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof/s 阅读全文
posted @ 2020-08-22 23:25 珊珊来迟0 阅读(118) 评论(0) 推荐(0) 编辑
摘要: solution naive class MinStack: def __init__(self): """ initialize your data structure here. """ self.stack = [] self.min_stack = [] def push(self, x: 阅读全文
posted @ 2020-08-16 23:05 珊珊来迟0 阅读(77) 评论(0) 推荐(0) 编辑
摘要: 看的题解, 叹为观止 class Solution: def spiralOrder(self, matrix: List[List[int]]) -> List[int]: ret = [] while matrix: ret += matrix.pop(0) matrix = list(zip( 阅读全文
posted @ 2020-08-16 22:45 珊珊来迟0 阅读(54) 评论(0) 推荐(0) 编辑
摘要: solution recursive class Solution: def isSymmetric(self, root: TreeNode) -> bool: def helper(q, p): if not q and not p: return True if not q and p: re 阅读全文
posted @ 2020-08-16 22:34 珊珊来迟0 阅读(55) 评论(0) 推荐(0) 编辑