import java.util.Arrays; /** * Created by ccc on 16-4-27. */ public class Test { public static void main(String arg[]) { , , }; ,,,, , }; Arrays.sort(a); Arrays.sort(b); int len = a.length; ; i < len; i++) { ) { System.out.println(b[i]); } } } } Arra…
目的:找出一个整形数组中的元素的最大值 以下,我们用类和对象的方法来做. #include<iostream> using namespace std; class Array_max{ private://声明在类的外部不可訪问的隐私成员 int array[10]; int max; public://声明在类的外部能够訪问的开放的成员函数 void set_value(){ int i; cout<<"请输入10个整数"<<endl;…
题目: There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 题意: 两个排序后的数组nums1 和nums2,长度分别是m,n,找出其中位数,并且时间复杂度:O(log(m+n)) 最愚蠢的方法: 两个数组合…
class Solution { public double findMedianSortedArrays(int[] A, int[] B) { int m = A.length; int n = B.length; if (m > n) { // to ensure m<=n int[] temp = A; A = B; B = temp; int tmp = m; m = n; n = tmp; } int iMin = 0, iMax = m, halfLen = (m + n + 1…
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. 这里题目要求找出出现次数超过n/2的元素. 可以先排序,…
