这样我们创建一个哈希表,对于每一个 x,我们首先查询哈希表中是否存在 target - x,然后将 x 插入到哈希表中,即可保证不会让 x 和自己匹配。 class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: hashtable = dict() for i, num in enumerat
deftwoSum(nums, target):iftarget / 2innums:#判断特殊情况l =[]foriinrange(len(nums)):ifnums[i] == target / 2: l.append(i)iflen(l) == 2:#还得判断是否得到了两个数,因为还有这种情况:[3, 2, 4] --> 6returnl d={}foriinrange(len(nums)): d[nums[i]]=iforiinrange(len(n...
小刀初试 [LeetCode] Two Sum, Solution Given an array of integers, 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. Please note ...
Can you solve this real interview question? Two Sum - Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may n
代码(java): 代码语言:txt AI代码解释 class Solution { public int[] twoSum(int[] numbers, int target) { int[] res = new int[2]; int i = 0, j = numbers.length - 1,temp;//i为左指针,j为右指针 while (i < j) { temp=numbers[i] + numbers[j];//先记录两数之和 ...
从栈中分别弹出栈顶数字 adder1 和 adder2,计算 adder1 和 adder2 之和,再加上进位 carry,得到当前位置的和 sum。 如果sum >= 10 ,那么进位 carry = 1...
代码实现-Java 01 解法一 由于加法需要从最低位开始运算,而最低位在链表末尾,链表只能从前往后遍历,没法取到前面的元素,那怎么办呢? 我们可以利用栈来保存所有的元素,然后利用栈的后进先出的特点就可以从后往前取数字了,我们首先遍历两个链表,将所有数字分别压入两个栈s1和s2中,我们建立一个值为0的head节点,然...
leetcode第28题--Divide Two Integers Divide two integers without using multiplication, division and mod operator. 分析:题目意思很容易理解,就是不用乘除法和模运 ... [LeetCode]29. Divide Two Integers两数相除 Given two integers dividend and divisor, divide two integers without using multiplication...
leetcode No1:https://leetcode.com/problems/two-sum/ Given an array of integers,returnindices of the two numbers such that they add uptoa specific target.You may assume that each input would have exactly one solution,andyou may not use the same element twice.Example:Given nums=[2,7,11,...
我的java code class Solution { public int solution(int[] A) { // write your code in Java SE 8 double avg_min = (A[0] + A[1])/2.0; int avg_min_pos = 0; double avg2, avg3; for(int i = 0; i < A.length - 2; i++) ...