这个代码完全不能通过LeetCode的测试。算法复杂度太高 然后我看了别人写的 1classSolution(object):2deftwoSum(self,nums,target):3"""4:type nums: List[int]5:type target: int6:rtype: List[int]7"""8n =len(nums)9result ={}10ifn <= 1:11returnFalse12else:13foriinrange(n):14ifnums[i]i...
result[0] =i result[1] =jreturnresult[:]//返回结果} } }returnnil } 回到顶部 四、C代码 int* twoSum(int* nums,intnumsSize,inttarget) {int*a = (int*)malloc(2*sizeof(int));for(inti =0;i < numsSize;i++){for(intj = i +1;j < numsSize;j++){if(nums[j] == target -nums...
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
因此我们可以用两个for循环遍历整个数组,找到这个数组中两个值的和等于这个给定值的数组下标并输出。 三、Go代码 AI检测代码解析 //1_常规解法 func twoSum(nums []int, target int) []int { var result = [2]int {0,0} if len(nums) < 2 { return nil } for i := 0 ; i < len(nums) - ...
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: hashtable = dict() for i, num in enumerate(nums): if target - num in hashtable: return [hashtable[target - num], i] hashtable[nums[i]] = i ...
这是每个初次接触leetcode的同学都将做的第一道题。题目本身的思维方式十分简单,可采用暴力破解法,利用for循环嵌套,便可通过测试: class Solution: def twoSum(nums: list, target: int) -> list: newlist = [] for firstIndex in range(0, len(nums)-1): for secondIndex in range(firstIndex+1, len...
[LintCode] Divide Two Integers 两数相除 Divide two integers without using multiplication, division and mod operator. If it is overflow, retu ... 029 Divide Two Integers 两数相除 不使用乘号,除号和取模符号将两数相除.如果溢出返回 MAX_INT.详见:https://leetcode.com/problems/divide-two-integer...
classSolution{ public: doublefindMedianSortedArrays(vector<int>&nums1,vector<int>&nums2) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2 nums1 = [1,3] nums2 = [2] 9 1 2 3 4 › [1,3] [2] [1,2] [3,4] ...
提交记录 提交记录 代码 题解不存在 请查看其他题解 9 1 2 3 4 5 6 › [2,1,4] [1,0,3] 5 [0,-10,10] [5,1,7,0,2] 18 Source 该题目是 Plus 会员专享题 感谢使用力扣!您需要升级为 Plus 会员来解锁该题目 升级Plus 会员
vector<int>twoSum(vector<int>&numbers,int target){vector<int>res;if(numbers.size()<2){returnres;}intleft=0,right=numbers.size()-1;while(left<right){if(target==numbers[left]+numbers[right]){res.push_back(left+1);res.push_back(right+1);returnres;}elseif(target>numbers[left]+numbers...