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
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...
这样我们创建一个哈希表,对于每一个 x,我们首先查询哈希表中是否存在 target - x,然后将 x 插入到哈希表中,即可保证不会让 x 和自己匹配。 class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: hashtable = dict() for i, num in enumerate(nums): if target - num ...
## 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...
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) ...
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...
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...
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; ...
2、Python解法 我是这样写的 classSolution(object):deftwoSum(self,nums,target):""":type nums: List[int] :type target: int :rtype: List[int]"""foriinrange(0,len(nums)-1):forjinrange(i+1,len(nums)):ifnums[i]+nums[j]==target:returni,j ...