1classSolution(object):2deftwoSum(self, nums, target):34forind, numinenumerate(nums):5iftarget-numinnumsandnums.index(target-num) !=ind:6return[ind, nums.index(target-num)]7return-1 直接利用了列表的索引。 运行速度仅为4ms 但是其中肯定用到了其他遍历比如nums.index(target-num)。这就是Pyt...
System.out.println(Arrays.toString(twoSum(new int[]{2,7,11,15}, 9))); System.out.println(Arrays.toString(twoSum(new int[]{3,2,4}, 6))); System.out.println(Arrays.toString(twoSum(newint[]{3,3}, 6))); } public static int[] twoSum(int[] nums, inttarget) { Map<Integer,In...
然后在备份的数组里找到位置。 1classSolution:2"""3@param numbers : An array of Integer4@param target : target = numbers[index1] + numbers[index2]5@return : [index1 + 1, index2 + 1] (index1 < index2)6"""7deftwoSum(self, numbers, target):8#write your code here9tmp =[]10for...
## 精简版 class Solution: def twosum2(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)): ## 遍历 nums,到i for j in range(i+1, len(nums)): ## 从 i 的右边寻找符合条件的元素 if nums[i] + nums[j] == target: return [i, j] ## 保存两个元...
Write a Python program to find the first two elements of a given list whose sum is equal to a given value. Use the itertools module to solve the problem. Sample Solution: Python Code: importitertoolsasitdefsum_pairs_list(nums,n):fornum2,num1inlist(it.combinations(nums[::-1],2))[:...
The solution is in the form of a Python dictionary. The dictionary keys are the variables and the dictionary values are the numerical solutions. We can access the solution out of the solution dictionary using regular dictionary indexing.
def solution(A): # write your code in Python 2.7 i, j, minSum = 0, len(A) - 1, 10 ** 10 A.sort() while i <= j: sum = A[i] + A[j] minSum = min(minSum, abs(sum)) if sum < 0: i += 1 elif sum > 0: j -= 1 else: return 0 return minSum Reply Sheng says...
nums = [2,7,11,15] & target = 9 -> [0,1], 2 + 7 = 9 At each num, calculate complement, if exists in hash map then return Time: O(n) Space: O(n) */ class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { int n = nums.size(); unordered_map...
int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode list = null, s = null; int sum = 0; while (l1 != null || l2 != null) { sum /= 10; if (l1 != null) { sum += ...
// } impl Solution { pub fn add_two_numbers_1( l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>, ) -> Option<Box<ListNode>> { let mut dump_head = ListNode::new(0); let mut current = &mut dump_head; let mut carry = 0...