1. Two Sum （easy）

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;
        }
    }
};