【python3/c++】leetcode 1. Two Sum (easy)
阿新 • • 發佈:2018-12-14
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the sameelement twice.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
1最優
2基礎思路:遍歷,但速度有點慢
class Solution: def twoSum(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ for i in range(len(nums)): if((target - nums[i]) in nums[i+1:]): return [i,nums[i+1:].index(target - nums[i])+i+1]
Runtime: 988 ms, faster than 27.56% of Python3
記錄一下看到的c++解法
class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, int> map; // key = desired number, value = paired index for (int i = 0; i < nums.size(); i++) { int val = nums[i]; // Try to find the current value in the hashmap. auto iter = map.find(val); if (iter != map.end()) { // Found! Return the previously-stored index and current index. return vector<int> {iter->second, i}; } // Didn't find an answer. Store the current index at "target - val". // That way we'll know that we have an answer as soon as we come across // the other part of the desired sum. E.g., if target == 12 and we // found a 7 here at index 2, store the index 2 in the map at key // 12 - 7 = 5. Then later when we find a "5" we're done. map[target - val] = i; } } };