[题目] Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue. Here, we will use the integers 0, 1, and 2 to represent the color red, white, a…

Sort a linked list in O(n log n) time using constant space complexity. 1.分析 该题主要考查了链接上的合并排序算法. 2.正确代码实现 package com.edu.leetcode; import com.edu.leetcode.ListNode; public class SortList { /** * @param args */ public ListNode Merge(ListNode first,List…

称号:Sort a linked list in O(n log n) time using constant space complexity. 思路:要求时间复杂度O(nlogn) 知识点:归并排序,链表找到中点的方法 存在的缺点:边界条件多考虑!.! /** * LeetCode Sort List Sort a linked list in O(n log n) time using constant space complexity. * 题目:将一个单链表进行排序,时间复杂度要求为o…

题目:Sort color <span style="font-size:18px;">/*LeetCode sort colors 题目:输入一个数组.包括0,1,2分别代表红白蓝三种颜色,要求依照0,1,2的顺序,将同类颜色的连续排列 思路:计数排序,是一个遍历两遍的方法:能够先统计每种的数量,之后直接将这一范围内的全部值都赋值为对应的数字就可以 遍历一遍的话能够在遍历的同一时候分别与0和2比較,从头和尾一起交换.1的在中间不用做处理: * */ package jav…

Sort Colors Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue. Here, we will use the integers 0, 1, and 2 to represent the color red, w…

Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue. Here, we will use the integers 0, 1, and 2 to represent the color red, white, and bl…

Sort a linked list in O(n log n) time using constant space complexity. 思路: 用归并排序.设输入链表为S,则先将其拆分为前半部分链表A,后半部分链表B.注意,要把A链表的末尾置为NULL.即要把链表划分为两个独立的部分,防止后面错乱. #include <iostream> #include <vector> #include <algorithm> #include <queue> #…

Sort List Sort a linked list in O(n log n) time using constant space complexity.   需要采用归并排序对链表进行操作.   归并排序思想:每次选取链表中间元素,把链表分割成两部分, 递归分割,直到链表中的元素是有序的时候(即只有一个元素时候),这时对链表进行合并, 合并的时候,每次选两个链表中小的元素放在下一个位置.     /** * Definition for singly-linked list. * str…

这个问题需要与归并排序排两个名单,基本思路分为切割与合并 合并后的代码Merge Two Sorted List里已经讲得非常清楚了. 所以这里直接给出代码. public ListNode merge(ListNode l1, ListNode l2) { ListNode helper = new ListNode(0); ListNode runner = helper; while (l1 != null && l2 != null) { if (l1.val < l2.va…

链表的归并排序 超时的代码 class Solution: def merge(self, head1, head2): if head1 == None: return head2 if head2 == None: return head1 # head1 and head2 point to the same link list if head1 == head2: return head1 head = None tail = None # the small pointer point…