zll00561的专栏

Write a function: class Solution { public int solution(int A, int B, int K); } that, given three integers A, B and K, returns the number of integers within the range [A..B] that are divisible by K, i

Determine whether given string of parentheses is properly nested. 需要注意空间复杂度为1，不能使用stack。 A string S consisting of N characters is called properly nested if: S is empty; S has the form “(U)” where U

You are given N counters, initially set to 0, and you have two possible operations on them: increase(X) − counter X is increased by 1, max counter − all counters are set to the maximum value of any c

A non-empty zero-indexed array A consisting of N integers is given. A permutation is a sequence containing each element from 1 to N once, and only once. For example, array A such that: A[0] = 4

A non-empty zero-indexed array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road. Array A contains only 0s and/or 1s: 0 represents a car tr

求一堵高低不平的墙，最少可以用多少块砖。砖可以是任意高度宽度的正方形。 You are going to build a stone wall. The wall should be straight and N meters long, and its thickness should be constant; however, it should have different heights

A non-empty zero-indexed array A consisting of N integers is given.A peak is an array element which is larger than its neighbors. More precisely, it is an index P such that 0 < P < N − 1, A[P − 1] < A

N块巧克力摆成一个环从0到N-1。首先吃No.0块，然后吃掉No.M块，依次吃掉No.2M….直到遇到空的块，求可以吃掉几块。 求最小公倍数的问题。class Solution { public int solution(int N, int M) { // write your code in Java SE 8 // x*M % N = y*M %N

// you can also use imports, for example: import java.util.*;// you can use System.out.println for debugging purposes, e.g. // System.out.println(“this is a debug message”);class Solution {

将某城市划分为m*n个区域，每个区域货物量保存在二维数组。将每个区的货物送到一个cargo center，汽车只能上，下，左，右，四个方向移动。求center使所有货物运到这儿的cost最小。初中物理有道题，给你一块不均匀木板，如何找到它的重心。用一根绳吊起来，画一条垂直线，再换个地方吊起了画出垂直线。两线交叉的地方就是重心。 这道题也是一样，先找出最小的列使所有点到这一列的cost最小。再找到行

(),[],{}配对的问题，使用stack的经典案例，不多说了。 强烈吐槽java泛型就是一坨屎。我又想换C#了，单就语言来说C#明显比java友好多了，后发优势吧。前浪都死在沙滩上了。最近的趋势是函数式编程语言。JS，D语言都有浓重的函数式味道class Solution { public int solution(String S) { // write your co

N voracious fish are moving along a river. Calculate how many fish are alive. You are given two non-empty zero-indexed arrays A and B consisting of N integers. Arrays A and B represent N voracious fis

Write a function: class Solution { public int solution(int[] A); } that, given a non-empty zero-indexed array A of N integers, returns the minimal positive integer that does not occur in A. For exam

求股票最大收益。算法导论上的问题 A zero-indexed array A consisting of N integers is given. It contains daily prices of a stock share for a period of N consecutive days. If a single share was bought on day P and sold

如果存在leader，返回其任意index，否则返回-1. A zero-indexed array A consisting of N integers is given. The dominator of array A is the value that occurs in more than half of the elements of A. For example, consider

求连续最大和子数组。与例子中不同的是，如果和为负返回最大的数而不是0.Solutution当maxCur+A[i]<0时，maxCur不能设置为0，因为如果所有值都小于0，得到错误结果。在之前程序的基础上，增加分段最大值为负的判断。class Solution { public int solution(int[] A) { int maxSum = Integer.MIN

求数组最小平均数子序列起始位置 A non-empty zero-indexed array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P < Q < N, is called a slice of array A (notice that the slice contains at

求两相邻子数组和最大，两端是开区间。 这道思索了很久，觉得解法没问题但是性能测试只有两个通过。 有两个因素影响最大值，两端的边界及切分的位置。边界扫描的同时在段内调整slice位置保证maxL+maxR的和最大。策略是以右边界扫描为主，maxR+A[i-1]<0及A[slice]>A[i-1]时调整slice。 但是为什么不对呢？？class Solution { public int

找出数组中出现不同元素的个数。import java.util.*;class Solution { public int solution(int[] A) { if(A.length==0) return 0; Arrays.sort(A); int size = 1; int former = 0; fo

Maximize A[P] * A[Q] * A[R] for any triplet (P, Q, R). 寻找数组A，三个数乘积的最大值。A non-empty zero-indexed array A consisting of N integers is given. The product of triplet (P, Q, R) equates to A[P] * A[Q] * A[R

Compute intersections between sequence of discs. 数组A[i]保存以i为圆点，A[i]为半径的圆。求多少对圆至少有一个共同点。 Given an array A of N integers, we draw N discs in a 2D plane such that the I-th disc is centered on (0,I) and

写的不错，代码看起来很舒服。欧几里德算法和扩展欧几里德算法 欧几里德算法 欧几里德算法又称辗转相除法，用于计算两个整数a,b的最大公约数。其计算原理依赖于下面的定理：定理：gcd(a,b) = gcd(b,a mod b)证明：a可以表示成a = kb + r，则r = a mod b 假设d是a,b的一个公约数，则有 d|a, d|b，而r = a - kb，因此d|r 因此