题目: 给定一个数组,求如果排序之后,相邻两数的最大差值,要求时间复杂度为O(N),且要求不能用非基于比较的排序 public static int maxGap(int nums[]) { if (nums == null || nums.length < 2) { return 0; } int len = nums.length; int max = Integer.MIN_VALUE; int min = Integer.MAX_VALUE; for (int i = 0; i < l
我们来看一个例子 Analysis Services: For Distinct Count measure NULL = 0 If you are to look at the table of values where you want to do a distinct count on the ID column ID ColA NULL blah NULL blah 0 blah 3 blah 1 blah 1 blah 2 blah 2 blah Within SQL, you wil
Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target. For example, given nums = [-2, 0, 1, 3], and target = 2. Retu
Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target. For example, given nums = [-2, 0, 1, 3], and target = 2. Retu
首先,我们在Model层中添加一个Person类. namespace MVCTest.Models{ public class Person { public string IdCard { get; set; } public string Name { get; set; } public string Sex { get; set; } public int Age { get; set; } }} 在Contro