题目来源：力扣（LeetCode）https://leetcode-cn.com/problems/two-sum 二、解答(java)：方案一：循环遍历，逐个尝试,时间复杂度为O(n^2)class Solution { public int[] twoSum(int[] nums, int target) { int a=0; int b=0; for(int i=0;i<nums.length;i++){ for(int j=i+...
class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { map<int, int> mapNums; vector<int> arr; int size = nums.size(); for(int i = 0; i < size; i ++) { map<int, int>::iterator iter = mapNums.find(target - nums[i]); if (iter != mapNums.end...
## 精简版 class Solution: def twosum2(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)): ## 遍历 nums,到i for j in range(i+1, len(nums)): ## 从 i 的右边寻找符合条件的元素 if nums[i] + nums[j] == target: return [i, j] ## 保存两个元...
1publicint[] twoSum(int[] nums,inttarget) {2Map<Integer, Integer> map =newHashMap<>();3for(inti = 0; i < nums.length; i++) {4intcomplement = target -nums[i];5if(map.containsKey(complement)) {6returnnewint[] { map.get(complement), i };7}8map.put(nums[i], i);9}10thr...
LeetCode-TwoSum问题 我的解法:借用HashMap,时间复杂度O(n),空间复杂度O(n) public int[] twoSum(int[] nums, int target) { Map<Integer,Integer> map = new HashMap<Integer,Integer>(); int i = 0; for(Integer num : nums){ map.put(num,i++);...
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 suchthat they add up to the target, where index1 must be less than index2.Please note that your returned answers (both index1 and...
5 这一步提供我的打败97%的人的代码实现代码：class Solution {public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int,int > map; for(int i=0;i<nums.size();i++) { int val=nums[i]; auto iter=map.find(val); if (iter!=map.end()) ...
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 return [] 官方给出的答案里,有些函数和语句可能不太了解,这里我说明一下...
public int[] twoSum(int[] nums, int target) { //遍历每个元素nums[i],查找是否存在值nums[j]相加后等于target。 for (int i = 0; i < nums.length; i++){ for(int j = i+1; j < nums.length;j++) { if (nums[i] + nums[j] == target) { ...
value < b.value; } class Solution { public: vector<int> twoSum(vector<int> &nums, int target) { int len = nums.size(); assert(len >= 2); vector<int> ret(2, 0); // 初始化:ret包含2个值为0的元素 vector<Node> nums2(len); for(int i = 0; i < len; i++){ nums2[i]...