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1. Two Sum【easy】
1. Two Sum【easy】 Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each in ...
分类:其他好文   时间:2017-09-17 17:35:32    阅读次数:130
[leetcode] 1. Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have ex ...
分类:其他好文   时间:2017-09-15 15:28:39    阅读次数:163
LeetCode——1. Two Sum
一.题目链接:https://leetcode.com/articles/two-sum/ 二.题目大意: 给定一个int型数组A和int值a,要求从A中找到两个数,使得这两个数值的和为a;返回结果为一个数组,该数组存储的为这两个数在数组A中的下标。(题目假设结果是唯一的) 三.题解 1.该题目首先 ...
分类:其他好文   时间:2017-09-15 12:22:08    阅读次数:115
[LeetCode] Two Sum IV - Input is a BST 两数之和之四 - 输入是二叉搜索树
Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target. E ...
分类:其他好文   时间:2017-09-12 09:51:05    阅读次数:345
Two Sum_LeetCode
Description: Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input ...
分类:其他好文   时间:2017-09-10 21:51:39    阅读次数:229
[LeetCode] 1 Two Sum
原题地址: https://leetcode.com/problems/two-sum/description/ 题目: Given an array of integers, return indices of the two numbers such that they add up to a ...
分类:其他好文   时间:2017-09-10 20:50:02    阅读次数:195
[LeetCode] Two Sum IV - Input is a BST
Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target. E ...
分类:其他好文   时间:2017-09-10 12:24:54    阅读次数:134
lintcode56 - Two Sum - easy
降低时间复杂度用的HashMap方法!不能用那个binarySearch,因为这里面要你返回的是index,binarysearch使用前要求你一定要sort过,不然他那个折半最后返回的值不会对的。而如果你sort过,那你返回的找到的那个index也和原始数字里所需数字的index不一样了! ...
分类:其他好文   时间:2017-09-09 15:51:06    阅读次数:123
Two Sum IV - Input is a BST
Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target. E ...
分类:其他好文   时间:2017-09-07 10:09:22    阅读次数:192
Two Sum IV - Input is a BST
这道题属于容易,还是挺简单的,但是我用的方法比较复杂。 题目: 思路: 1、我的想法是把树转换成列表,然后再遍历这个列表,判断是否两个元素相加为0,这种方法时间复杂度较高 2、大神也是将树节点存在列表中,建立一个空集合set,遍历列表,如果该(k - 该节点值)在集合中则返回True,否则将该节点值 ...
分类:其他好文   时间:2017-09-06 13:01:43    阅读次数:194
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