vector<int> twoSum2(const vector<int> &numbers, int target) { vector<int> result; if(numbers.empty()) { return result; } unordered_map<int, int> num2loc; //space: O(N) for(int i = ; i < numbers.size(); ++i) //哈希映射,time: O(N) { num2loc[ numbers[i] ] = i; }...
Given an array of integers that is alreadysorted in ascending order, 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 ...
//O(nlgn)classSolution {public: vector<int> twoSum(vector<int>& numbers,inttarget) {for(inti =0; i < numbers.size(); ++i) {intt = target - numbers[i], left = i +1, right = numbers.size();while(left <right) {intmid = left + (right - left) /2;if(numbers[mid] == t)...
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。 这里我只写了2Sum和3Sum的代码,注意要避免重复排序,同时避...
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...
twoSum 方法(2):哈希法。只是求两个值之和,那么知道其中一个值,另一个如果存在于序列中,那么就找到了。这里用unordered_map实现,将所有元素装进该unordered_map中,元素值作为key,位置作为元素。接着,对每个序列中的元素,用map的find看是否有另一伴的存在,若不存在就掠过,找到还得小心个坑,也就是找到的另一伴...
3 输入与输出：vector<int> twoSum(vector<int>& nums, int target){}完成这个成员函数解决问题。4 思路：这个可以使用哈希表一次搞定这个问题。当我们扫描整个数组，检查当前元素的补码是否已经存在于表中。如果存在，我们已经找到解决方案并立即返回。如果不存在就向表中插入该元素。5 这一步提供我的打败97%的人...
2 3 4 5 6 classSolution{ public: vector<int>twoSum(vector<int>&nums,inttarget) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2Case 3 nums = [2,7,11,15] target = 9 9 1 2 3 4 5 6 › [2,7,11,15] ...
Given an array of integers that is already sorted in ascending order, 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...
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2. 1.2 中文描述 给你一个整数数组,并且这个数组是按递增排序的,你要找到数组中的两个整数,它们的和等于给定的目标值,然后返回它们的下标。 题目假设给你的数组总是有且只有一个解,而且同一个元素不能使用两次。另外,返回结果的...