刷题记录

Problem: Compare two version numbers  version1  and  version2 . If  version1  >  version2  return  1;  if  version1  <  version2  return  -1; otherwise return  0 . You may assume that the version strings are non-empty and contain only digits and the  .  character. The  .  character does not represent a decimal point and is used to separate number sequences. For instance,  2.5  is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision. Example 1: Input: version1 = "0.1", version2 = "1.1" Output: -1 Example 2: Input: version1 = "1.0.1", version2 = "1" Output: 1 Example 3: Input: version1 = "7.5.2.4", version2 = "7.5.3" Output: -1 Analysis: 这道题实现的时候要注意在数组越界的时候把值赋为零。 Solution: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 class Solution { public int compareVersion ( String version1 , Str

Problem: Given a singly linked list, return a random node's value from the linked list. Each node must have the  same probability  of being chosen. Follow up: What if the linked list is extremely large and its length is unknown to you? Could you solve this efficiently without using extra space? Example: // Init a singly linked list [1,2,3]. ListNode head = new ListNode(1); head.next = new ListNode(2); head.next.next = new ListNode(3); Solution solution = new Solution(head); // getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning. solution.getRandom(); Analysis: 实在是看不懂reservoir sampling, 看了2个小时稍微有点儿眉目。。。意思就是随机取一个范围1到i的数，可以保证概率随机。。。。 Solution: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class