k:int)->int:# 将问题转化为寻找第n-k个最小元素k=len(nums)-kdefquickSelect(l,r):pivot,p=nums[r],l# 将小于等于pivot的元素移动到左侧foriinrange(l,r):ifnums[i]<=pivot:nums[p],nums[i]=nums[i],nums[p]p+=1# 将pivot放到正确的位置上nums[p],nums[r]=nums[r],nums[p]# 如果p...
What is the largest axis-aligned plus sign of 1s contained in the grid? Return the order of the plus sign. If there is none, return 0. An “axis-aligned plus sign of 1s of order k” has some center grid[x][y] = 1 along with 4 arms of length k-1 going up, down, left, ...
代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution { public: std::vector<int> largestDivisibleSubset(std::vector<int>& nums) { std::sort(nums.begin(), nums.end()); auto dp = std::vector<int>(nums.size(), 0); auto parent = std::vector<int>(nums.size(), 0); in...
In this program, only the if statement is executed when n1 is the largest.Similarly, only the else if statement is executed when n2 is the largest, and so on.Example 3: Using Nested if...else#include <stdio.h> int main() { double n1, n2, n3; printf("Enter three numbers: "); ...
class Solution { /* * @param k : description of k * @param nums : array of nums * @return: description of return */ public int kthLargestElement(int k, int[] nums) { if (nums == null || nums.length == 0 || k < 1 || k > nums.length){ return -1; } return partition(...
packagecom.includehelp.basicimport java.util.*// Main Method Entry Point of Programfunmain(args: Array<String>) {// InputStream to get Inputvalreader = Scanner(System.`in`)// Input Integer valuesprintln("Enter First Value : ")varfirst = reader.nextInt() println("Enter Second Value : "...
util.*; public class LargestNumber{ public static void main(String []args) { int a=0,b=0,c=0; int largest=0; //Scanner class to take user input. Scanner X = new Scanner(System.in); System.out.print("Enter First No. :"); a = X.nextInt(); //read integer number System.out...
import Foundation import Glibc func maximumValue(n1: Int, n2: Int, n3: Int) -> Int{ // Comparing n1, n2 and n3 with // each other to find the largest number if n1 > n2, n1 > n3{ return n1 } else if n2 > n3, n2 > n1{ return n2 } else if n3 > n2, n3 > n1{ return...
intstart=stack.isEmpty() ? -1: stack.peekLast(); intarea=height * (i - start -1); res = Math.max(res, area); } stack.offerLast(i); } returnres; } } JavaScript实现 /** *@param{number[]}heights *@return{number} */
classSolution{public:intlargestRectangleArea(vector<int>&heights){stack<int>h;//只需存储下标,高度可从heights中直接读出heights.push_back(0);//用于遍历完heights中所给元素后,清空栈intans=0,hsize=heights.size();for(inti=0;i<hsize;i++){while(!h.empty()&&heights[h.top()]>heights[i]){int...