这个是LeetCode鼻祖 两数之和 题目的升级版本,不过难度也只是略有提升。做法和之前的 Two Sum 差别不大。 回顾下两数之和 题目1(Two Sum): 问题描述:给定一个整数数组 nums 和一个目标值 target,请你在该数组中找出和为目标值的那两个整数,并返回它们的数组下标。 条件:数组未排序,每种输入只会对应一个答...
Can you solve this real interview question? Two Sum II - Input Array Is Sorted - Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two
1.1 英文题目 Given an array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Return the indices of the two numbers (1-indexed) as an integer array answer of size 2, where 1 <= answer[0] < answer[...
Given an array of integers that is alreadysorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Note: Your ...
Given an array nums, return true if the array was originally sorted in non-decreasing order, then rotated some number of positions (including zero). Otherwise, return false. There may be duplicates in the original array. Note: An array A rotated by x positions results in an array B of ...
Leetcode学习(2)—— Two Sum II - Input array is sorted,Givenanarrayofintegersthatisalreadysortediner.ThefunctiontwoSumshouldreturnindicesofthetwonumberssuc
原题链接在这里:https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/ 题目: Given an array of integers that is alreadysorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such...
Golang Leetcode 167. Two Sum II - Input array is sorted.go 发布于2019-04-12 11:22:36 41700 代码可运行 举报 文章被收录于专栏:学习日记 关联问题 换一批 Golang中如何实现Two Sum II算法? 在Golang中如何处理已排序数组的两数之和问题? Golang Leetcode 167题目的时间复杂度是多少? 版权声明:原...
if (temp == target)先判断与目标数是否相同 可减少运行时间(因为Leetcode是拿很多不同数据来运行,而不是一条超长数据。仔细想想这里的区别) temp=numbers[i] + numbers[j]先把两数之和记录下来,像py3里那种判断两次(==、>)每次都计算一次两数和,会消耗更多时间,这在判断条件增多时会很明显。
if (temp == target)先判断与目标数是否相同 可减少运行时间(因为Leetcode是拿很多不同数据来运行,而不是一条超长数据。仔细想想这里的区别) temp=numbers[i] + numbers[j]先把两数之和记录下来,像py3里那种判断两次(==、>)每次都计算一次两数和,会消耗更多时间,这在判断条件增多时会很明显。