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)...
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...
sum -= candidates[i]; path.pop_back(); } 注意sum + candidates[i] <= target为剪枝操作,在39.组合总和有讲解过! C++代码 回溯三部曲分析完了,整体C++代码如下: class Solution { private: vector<vector<int>> result; vector<int> path; void backtracking(vector<int>& candidates, int target, int...
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...
You may assume that each input would have exactly one solution. Input:numbers={2, 7, 11, 15}, target=9 Output:index1=1, index2=2 » Solve this problem [Thoughts] 两种解法。 解法一, hash 从左往右扫描一遍,然后将数及坐标,存到map中。然后再扫描一遍即可。时间复杂度O(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。 回到顶部 三、Go代码 //1_常规解法func twoSum(nums []int, targetint) []int{varresult = [2]int{0,0}iflen(nums) <2{returnnil }fori :=0; i < len(nums) -1; i++{forj := i +1; j...
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} ...
, 1-based) as value. Then for eachnumofnums, search fortarget - numin the hash table. If it is found and is not the same asnum, then we are done. Notice that the problem statement has excluded the case of duplicates by stating that "each input would have exactly one solution"....
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 <...
class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode head = null, tail = null; int carry = 0; while (l1 != null || l2 != null) { int n1 = l1 != null ? l1.val : 0; int n2 = l2 != null ? l2.val : 0; int sum = n1 + n2 + carry; ...