Given an array of integers, the majority number is the number that occurs more than half of the size of the array. Find it. Notice You may assume that the array is non-empty and the majority number always exist in the array. Have you met this questio

Yet Another Source Code for LintCode Current Status : 232AC / 289ALL in Language C++, Up to date (2016-02-10) For more problems and solutions, you can see my LintCode repository. I'll keep updating for full summary and better solutions. See cnblogs t

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. Hint: How many majority elements could it possibly have? Do you have a better hint? Suggest it! 这道题让我们

--------------------------------------------------------------- 本文使用方法:所有题目,只需要把标题输入lintcode就能找到.主要是简单的剖析思路以及不能bug-free的具体细节原因. ---------------------------------------------------------------- ------------------------------------------- 第九周:图和搜索. ---

-------------------------------------------- AC代码: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ pub

----------------------------------- Moore's voting algorithm算法:从一个集合中找出出现次数半数以上的元素,每次从集合中去掉一对不同的数,当剩下一个元素的时候(事实上只要满足一个元素出现过半就一定会剩下一个元素的)这个元素就是我们要找的数了. AC代码: public class Solution { /** * @param nums: a list of integers * @return: find a majority numb

----------------------------------- 最开始的想法是先计算出链表的长度length,然后再从头走 length-n 步即是需要的位置了. AC代码: /** * Definition for ListNode. * public class ListNode { * int val; * ListNode next; * ListNode(int val) { * this.val = val; * this.next = null; * } * } */ pu

------------------------ 因为字符究竟是什么样的无法确定(比如编码之类的),恐怕是没办法假设使用多大空间(位.数组)来标记出现次数的,集合应该可以但感觉会严重拖慢速度... 还是只做出了O(n^2)... 勉强AC代码: public class Solution { /** * @param str: a string * @return: a boolean */ public boolean isUnique(String s) { for(int i=0;i<s.

-------------------- 递归那么好为什么不用递归啊...我才不会被你骗...(其实是因为用惯了递归啰嗦的循环反倒不会写了...o(╯□╰)o) AC代码: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left

----------------------------------- AC代码: /** * Definition for ListNode. * public class ListNode { * int val; * ListNode next; * ListNode(int val) { * this.val = val; * this.next = null; * } * } */ public class Solution { /** * @param node: the node

----------------------------------------------- AC代码: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */

-------------------------- 水题 AC代码: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solut

------------------------ 只要设置两个指针,称为快慢指针,当链表没有环的时候快指针会走到null,当链表有环的时候快指针早晚会追上慢指针的. AC代码: /** * Definition for ListNode. * public class ListNode { * int val; * ListNode next; * ListNode(int val) { * this.val = val; * this.next = null; * } * } */ publi

--------------------------------------- 按照给定的峰值定义,峰值的左半部分一定是递增的,所以只要找到不递增的即可. AC代码: class Solution { /** * @param A: An integers array. * @return: return any of peek positions. */ public int findPeak(int[] A) { for(int i=1;i<A.length;i++){ if(A[i]>=

------------------------------------------------------------ 卧槽竟然连题意都没看懂,百度了才明白题目在说啥....我好方啊....o(╯□╰)o AC代码: class Solution { /** * @param prices: Given an integer array * @return: Maximum profit */ public int maxProfit(int[] prices) { int res=0; fo

-------------------------------- 这道题好坑啊,自己说是2*n+1个数字,结果有组测试数据竟然传了个空数组进来... 经典位算法: n^n==0 n^0==n AC代码: public class Solution { /** *@param A : an integer array *return : a integer */ public int singleNumber(int[] A) { if(A.length==0) return 0; int n=A

--------------------------------- AC代码: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class S

------------------------ AC代码: class Solution { /** * param n: As description. * return: A list of strings. */ public ArrayList<String> fizzBuzz(int n) { ArrayList<String> results = new ArrayList<String>(); for (int i = 1; i <= n; i++

木头加工 题目描述 有一些原木,现在想把这些木头切割成一些长度相同的小段木头,需要得到的小段的数目至少为 k.当然,我们希望得到的小段越长越好,你需要计算能够得到的小段木头的最大长度. 注意事项 木头长度的单位是厘米.原木的长度都是正整数,我们要求切割得到的小段木头的长度也要求是整数.无法切出要求至少 k 段的,则返回 0 即可. 样例 有3根木头[232, 124, 456], k=7, 最大长度为114,则返回114. 简单分析 暴力解法就是从1开始遍历所有的长度,对于长度L,计算所有木头可

题目1 落单的数 给出2*n + 1 个的数字,除其中一个数字之外其他每个数字均出现两次,找到这个数字. 链接:http://www.lintcode.com/zh-cn/problem/single-number/ 样例 给出 [1,2,2,1,3,4,3],返回 4 挑战 一次遍历,常数级的额外空间复杂度 解决方案 方法1思路:将所有的数转换成二进制,因为是int类型,共32位.申请常数级(32位)的额外空间,然后每个数对应的位相加,最后对应位上的和模2.最后的结果就是单个数对应的二进制数.