码迷,mamicode.com
首页 >  
搜索关键字:spoj balnum balanced    ( 1777个结果
[Leetcode][Tree][Balanced Binary Tree]
判断一棵树是不是平衡二叉树,之前做过,还有点印象,用一个函数返回树的高度,如果是-1的话,就说明子树不平衡。1A很开心~ 1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * T...
分类:其他好文   时间:2014-07-07 13:56:05    阅读次数:180
Problem Balanced Binary Tree
Problem Description:Given a binary tree, determine if it is height-balanced.For this problem, a height-balanced binary tree is defined as a binary tre...
分类:其他好文   时间:2014-07-07 13:27:35    阅读次数:207
[LeetCode]Balanced Binary Tree
[LeetCode]Balanced Binary Tree...
分类:其他好文   时间:2014-07-04 07:54:30    阅读次数:199
[leetcode] Convert Sorted List to Binary Search Tree
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
分类:其他好文   时间:2014-07-03 19:10:40    阅读次数:201
[leetcode] Balanced Binary Tree
Given a binary tree, determine if it is height-balanced.
分类:其他好文   时间:2014-07-03 19:00:17    阅读次数:188
[leetcode] Convert Sorted Array to Binary Search Tree
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
分类:其他好文   时间:2014-07-03 18:55:48    阅读次数:236
SPOJ 220 Relevant Phrases of Annihilation (后缀数组)
题目大意: 求在m个串中同时出现两次以上且不覆盖的子串的长度。 思路分析: 二分答案,然后check是否满足,判断不覆盖的方法就是用up down 来处理边界。 #include #include #include #include #include #include #define maxn 110005 using namespace std; char ...
分类:其他好文   时间:2014-07-03 16:55:30    阅读次数:350
SPOJ 8222 NSUBSTR Substrings
Substrings Time Limit: 1000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status]   Description You are given a string S which cons...
分类:其他好文   时间:2014-07-03 16:52:59    阅读次数:166
spoj 694 求一个字符串中不同子串的个数
SPOJ Problem Set (classical) 694. Distinct Substrings Problem code: DISUBSTR Given a string, we need to find the total number of its distinct substrings. Input T- number of ...
分类:其他好文   时间:2014-07-03 16:32:42    阅读次数:212
SPOJ 694、705 Distinct Substrings 、 New Distinct Substrings (后缀数组)
题目大意: 求串中不同的子串的个数。 思路分析: 子串一定是某一个后缀的前缀。 所以我们把每一个后缀拿出来,分析它有多少个前缀,然后除去它与sa数组中前面那个后缀相同的前缀。 最后也就是 ans = segma (n-sa[i] + height[i]).... #include #include #include #include #define maxn 1...
分类:其他好文   时间:2014-07-02 07:23:21    阅读次数:182
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!