1.Two Sum

1.Two Sum

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.

Example:

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

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

雙循環最簡單的實現：

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

            }
        }

        //no result
        result.push_back(-1);
        result.push_back(-1);
        return result;

    }
};

採用hash表：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        // hash[i]表示nums中數值爲i的下標
        unordered_map<int, int> hash;
        vector<int> result;

        // 一邊循環每個數，一邊加入hash表。
        for (int i = 0; i < nums.size(); i++) {
            if (hash.find(target - nums[i]) != hash.end()) {
                // target - nums[i]的下標更小，放在前面
                result.push_back(hash[target - nums[i]]);
                result.push_back(i);
                return result;
            }
            hash[nums[i]] = i;
        }

        // 無解的情況
        result.push_back(-1);
        result.push_back(-1);
        return result;
    }
};