Console.ReadKey(); }publicint[] TwoSum(int[] nums,inttarget) {vardictionary=newDictionary<int,int>();for(inti =0; i < nums.Length; i++) {varnn = target -nums[i];if(dictionary.ContainsKey(nn)) {returnnewint[]{ dictionary[nn],i}; } dictionary[nums[i]]=i; }thrownewArgumentExcep...
四、代码 1/*2* 新建结构体,记录nums中每个元素的下标和数据,以便排序处理3*/4typedefstructObject5{6intindex;7intvalue;8} Object;910/*11* 用于 qsort12*/13intcompare(constvoid*objA,constvoid*objB)14{15return(((Object *)objA)->value - ((Object *)objB)->value);16}1718int* twoSum(int...
如何用PYTHON找出Two Sum的结果 工具/原料 PYTHON 方法/步骤 1 新建一个PY文档，打开JUPTER NOTEBOOK。2 #Given nums = [2, 7, 11, 15], target = 9nums = [2, 7, 11, 15]target = 9我们要找出列表里面两个相加数为目标的数字。3 nums = [2, 7, 11, 15]target = 9nums2 = numsfor a, ...
vector<vector<int>> fourSum(vector<int>& num,inttarget) { vector<vector<int>>result;if(num.size() <4)returnresult; sort(num.begin(), num.end()); auto last=num.end();for(auto a = num.begin(); a < prev(last,3); ++a) {for(auto b = next(a); b < prev(last,2); ++b)...
vector<int> twoSum(vector<int>& nums, int target) {vector<int> result; std::multimap <int, int> value; std::vector<int>::const_iterator cit = nums.begin();int index =0; for (; cit != nums.end() ; index++, cit++) {
针对2Sum,先对数组排序,然后使用双指针匹配可行解就可以解决,虽然可以考虑使用HashMap加速搜索,但是对于本题使用HashMap的与否的时间复杂度都一样,都是O(nlog(n))。可以参考这个链接: 求和问题总结(leetcode 2Sum, 3Sum, 4Sum, K Sum),写的很清楚。
如何解决Leetcode的Two Sum问题? Two Sum问题的时间复杂度是多少? Question: 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...
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It may save you a few keystrokes in a small program like this one, but can turn into a major maintenance headache for a codebase that you need to support. See this question. long long int n,x,a,b,sum=0,s=1; Never declare multiple variables on a single line. It's very easy ...
classSolution{public:vector<int>twoSum(vector<int>&nums,inttarget){vector<int>result={0,0};intlength=nums.size();for(inti=0;i<length;i++){for(intj=i+1;j<length;j++){if(nums[i]+nums[j]==target){result[0]=i;result[1]=j;}}}returnresult;}}; ...