Leecode Week1: Two Sum

Week1：Two Sum

1. Question

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 same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

2. Solution

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        map<int,int> b;
        for(int i = 0; i < nums.size(); i++) {
            b.insert(pair<int,int>(nums[i], i));
        }

        for(int i = 0; i < nums.size(); i++) {
            int temp = target - nums[i];
            if(b.count(temp) && b[temp] != i) {
                return vector<int>{i, b[temp]};
            }
        }
        return vector<int>{};
    }
};

思路:利用map將數組元素和下標對應起來，然後用一個循環來得到每個元素所要匹配的值並在map中查找是否存在這個值，若存在就返回兩個元素的下標，否則繼續查找。這裏利用map降低了搜索的複雜度，時間複雜度爲O(nlog(n))。需要注意的是，每個元素只能使用一次，所以需要判斷是否匹配的是同一個值。