方法2 AC代码: class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> res(2); // 双指针, 先固定一个 for (int i = 0; i < nums.size(); i++) { for (int j = i + 1; j < nums.size(); j++) { if (nums[i] + nums[j] == target)...
## 哈希表解法,精简版 class Solution: def twosumhx2(self, nums: List[int], target: int) -> List[int]: values = {} for i, num in enumerate(nums): ## i=index, num=num diff = target - num if diff in values: return [values[diff], i] ## 返回的是2个索引 values[num] = i...
else if (sum < 0) //和太小,p向后移动 { ++p; } else //和过大,q向前移动 { --q; } } } <span style="font-family: Arial, Helvetica, sans-serif;">// 3 sum cloest class Solution {</span> public: int threeSumClosest(vector<int> &num, int target) { int index; bool flag=t...
解法一实现如下: 1: vector<int>twoSum(vector<int> &numbers, int target) {2: map<int, int> mapping;3: vector<int> result;4:for(int i =0;i< numbers.size();i++)5: {6: mapping[numbers[i]]=i;7: }8:for(int i =0;i< numbers.size();i++)9: {10: int searched = target -...
leetcode-cn.com/problem You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list. You may assume the two numbers do not...
思路2:考虑使用map或者hash_map,遍历输入的numbers,在map映射表中寻找target-numbers[i],C++ STL中map使用红黑树实现,查找元素时间O(logn),总时间O(nlogn) 1classSolution {2public:3vector<int> twoSum(vector<int> &numbers,inttarget) {45vector<int>result;6map<int,int>hashMap;78for(inti =0; i <...
这几天在做LeetCode 里面有2sum, 3sum(closest), 4sum等问题, 这类问题是典型的递归思路解题。该这类问题的关键在于,在进行求和求解前,要先排序Arrays.sort()可实现,而本文则着重探讨关于KSum问题。 leetcode求和问题描写叙述(K sum problem): K sum的求和问题通常是这样子描写叙述的:给你一组N个数字(比方...
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 ...
求和问题描述(K Sum problem): 给你一组N个数字(比如 vector num), 然后给你一个目标常数(比如 int target) ,我们的目的是在这一堆数里面找到K个数字,使得这K个数字的和等于target。 K Sum求解方法, 适用2Sum, 3Sum, 4Sum: 方法一:暴力,就是枚举所有的K-subset, 那么这样的复杂度就是 从N选出K个,复...
C++ Solution: 1vector<int> twoSum(vector<int>& nums,inttarget) {2unordered_map<int,int>_map;3for(inti =0; i < nums.size(); i++)4{5inttemp = target -nums[i];6if(_map.find(temp) !=_map.end()) {7return{ _map[temp], i};8}9_map[nums[i]] =i;10}11return{};12} ...