摘要： 最近遇到了这么一个需求，要接一家外部供应商，其中牵涉到接口回调，对方会把一些信息通过调用我们的接口发送给我们。 那么问题来了： 1. 我想在本地调试 2. 本地是内网环境 3. 公司的ip也不是固定的(==！) 4. 我手里恰好有一台便宜的阿里云机器，固定ip 所以最方便的做法当然是阿里云和我本地机 阅读全文

摘要： The algorithm for this problem is not so hard to figure out. This is a problem of finding the length of the shortest path in graph. As I mentioned in 阅读全文

摘要： ```java class Solution { public int knightDialer(int N) { int cons = 1000000007; int[][] d = new int[10][N]; int[][] m = new int[][]{{4, 6, 1}, {6, 8, 阅读全文

摘要： S has a maximum length of 1000, and it's easy to figure out the total count of numbers combined by substring of S. Which is 1 + min(2, S) + min(2^2, S 阅读全文

摘要： For each position, use A[i] to record the max value of A[j] + j, which j 阅读全文

摘要： ```java class Solution { public boolean canThreePartsEqualSum(int[] A) { int sum = 0; for (int i = 0; i 阅读全文

摘要： Before the first AC, I was not quite sure if Arrays.sort is stable. so I wrote something like this. After read Arrays.sort's doc. (Surely stable sort) 阅读全文

摘要： ```java class Solution { public int shipWithinDays(int[] weights, int D) { int left = 1, right = 50000; int N = weights.length; for (int i = 0; i cap) { co... 阅读全文

摘要： time =1 and = 2 and sum 500) continue; if (counter[target] != 0) ret += counter[target]; } counter[time[i]] += 1; } return ret; } } java public int nu 阅读全文

摘要： Suppose the given number is N, the result is R, it's easy to notice that N + R + 1 = 2^x, and x is unknown, but it's quite easy to calculate. for x fr 阅读全文