Sort K-Sorted Array: You are given an array of n elements such that each element is at max k place away from its original sorted position, you have to sort the array in correct manner. Submitted byDivyansh Jaipu
sort(array); System.out.println("Sorted array: " + Arrays.toString(array)); } } 输出: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 Sorted array: [apple, banana, grape, orange] 示例三:对自定义对象数组排序 代码语言:javascript 代码运行次数:0 运行 AI代码解释 import java.util.Arrays; ...
Given an array of n elements, where each element is at most k away from its target position, you need to sort the array optimally. Example 1: Input: n = 7, k = 3 arr[] = {6,5,3,2,8,10,9} Output: 2 3 5 6 8 9 10 Explanation: The sorted array will be 2 3 5 6 8 9...
* Sorts the specified range of the array by Dual-Pivot Quicksort. * * @param a the array to be sorted * @param left the index of the first element, inclusive, to be sorted * @param right the index of the last element, inclusive, to be sorted * @param leftmost indicates if this p...
Functions-in-k Functions-l Functions-m-r Functions-s Functions-t-z ColdFusion tags CFML Reference Cloud services DescriptionSorts array elements numerically or alphanumerically.ReturnsTrue, if sort is successful; False, otherwise. The member function returns the sorted array....
num_list=[1,8,2,3,10,4,5]ordered_list=sorted(range(len(num_list)),key=lambdak:num_list[k])print(ordered_list)# [0, 2, 3, 5, 6, 1, 4] 4.字符串类型排序 # 字符串类型排序str_list=['1', '8', '2', '3', '10', '4', '5']ordered_list=sorted(str_list)print(ordered...
int[] run = new int[MAX_RUN_COUNT + 1]; int count = 0; run[0] = left; // Check if the array is nearly sorted for (int k = left; k < right; run[count] = k) { if (a[k] < a[k + 1]) { // ascending while (++k <= right && a[k - 1] <= a[k]); } else...
// Checkifthe array is nearly sortedfor(int k = left; k < right; run[count] = k) {if(a[k] < a[k + 1]) { // ascendingwhile(++k <= right && a[k - 1] <= a[k]); }elseif(a[k] > a[k + 1]) { // descendingwhile(++k <= right && a[k - 1] >= a[k]);fo...
return; // Arrays of size 0 and 1 are always sorted //如果数组长度小于MIN_MERGE(32)则使用二分排序 // If array is small, do a "mini-TimSort" with no merges if (nRemaining < MIN_MERGE) { int initRunLen = countRunAndMakeAscending(a, lo, hi, c); ...
int a1 = a[k], a2 = a[left]; //a1始终存最大值,如果不是,则swap,确保a1最大 if (a1 < a2) { a2 = a1; a1 = a[left]; } //先插入最大值 while (a1 < a[--k]) { a[k + 2] = a[k]; } a[++k + 1] = a1;