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摘要: 题目描述: class Solution: def checkOverlap(self, radius: int, x_center: int, y_center: int, x1: int, y1: int, x2: int, y2: int) -> bool: a = max(0, x1 - x 阅读全文
posted @ 2020-04-05 10:23 oldby 阅读(222) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 恢复内容开始 恢复内容开始 选择排序 O(N^2) class Solution { public int[] sortArray(int[] nums) { int len = nums.length; for(int i = 0; i < len -1; i++){ int minI 阅读全文
posted @ 2020-04-03 11:54 oldby 阅读(106) 评论(0) 推荐(0) 编辑
摘要: 恢复内容开始 选择排序 O(N^2) class Solution { public int[] sortArray(int[] nums) { int len = nums.length; for(int i = 0; i < len -1; i++){ int minIndex = i; for 阅读全文
posted @ 2020-04-03 09:32 oldby 阅读(123) 评论(0) 推荐(0) 编辑
摘要: 题目描述: java /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class 阅读全文
posted @ 2020-04-02 17:06 oldby 阅读(114) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一:java class Solution { public String replaceSpace(String s) { StringBuilder res = new StringBuilder(); for(Character c: s.toCharArray()){ if( 阅读全文
posted @ 2020-04-02 16:55 oldby 阅读(159) 评论(0) 推荐(0) 编辑
摘要: 解法一:O(MN)O(MN))Java版 class Solution { public void gameOfLife(int[][] board) { int [] neighbors = {0,1,-1}; int rows = board.length; int cols = board[0 阅读全文
posted @ 2020-04-02 16:45 oldby 阅读(178) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一:pythonO(N)O(1) class Solution: def maxDepthAfterSplit(self, seq: str) -> List[int]: d = 0 ans = [] for s in seq: if s == "(": d += 1 ans.app 阅读全文
posted @ 2020-04-01 12:13 oldby 阅读(184) 评论(0) 推荐(0) 编辑
摘要: 恢复内容开始 题目描述:给定一个数组,返回排序后的数组。 python 归并排序: class Solution: def sortArray(self, nums: List[int]) -> List[int]: def merge_sort(nums): if len(nums) <= 1: 阅读全文
posted @ 2020-04-01 11:05 oldby 阅读(164) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一:第一次提交:python暴力 O(N^2) class Solution: def lastRemaining(self, n: int, m: int) -> int: a = [i for i in range(n)] i = 0 while(len(a) > 1): i = 阅读全文
posted @ 2020-03-31 10:33 oldby 阅读(124) 评论(0) 推荐(0) 编辑
摘要: 题目描述: python 回溯 +kmp next数组 class Solution: def findGoodStrings(self, n: int, s1: str, s2: str, evil: str) -> int: MOD = int(1e9 + 7) m = len(evil) or 阅读全文
posted @ 2020-03-30 16:15 oldby 阅读(200) 评论(0) 推荐(0) 编辑
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