Leetcode 1. Two Sum (C++ & Python)

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].

這題不難，首先想到的就是暴力遍歷，用兩個for循環就能做出來，但是時間太長

//Solution1
class Solution {
public:
	vector<int> twoSum(vector<int>& nums, int target) {
		vector<int> result;
		int i, j;
		for (i = 0; i < nums.size(); i++) {
			for (j = i + 1; j < nums.size(); j++) {
				if (nums[i] + nums[j] == target) {
					result.emplace_back(i);
					result.emplace_back(j);
                    return result;
				}
			}
		}
		return result;
    }
};

後來也是看了其他人的解法，發現用HashMap的思想會比較快。於是用了Python的字典dict{}

#Solution2
def twoSum(self, nums, target):
    """
    :type nums: List[int]
    :type target: int
    :rtype: List[int]
    """
    n = len(nums)
    dic = {}
    for x in range(n):
        a = target - nums[x]
        if nums[x] in dic:          #關鍵步驟
            return dic[nums[x]], x
        else:
            dic[a] = x;

實例步驟如下：