LeetCode 1. Two Sum

原题链接在这里：https://leetcode.com/problems/two-sum/

题目：

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].

题解：

采用HashMap记录之前出现的num 和 其对应的 index.

Time Complexity: O(n). Space: O(n).

AC Java:

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        if(nums == null || nums.length < 2){
            throw new IllegalArgumentException("Invalid input.");
        }
        
        int [] res = {-1, -1};
        HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
        for(int i = 0; i<nums.length; i++){
            if(!hm.containsKey(target - nums[i])){
                hm.put(nums[i], i);
            }else{
                res[0] = hm.get(target-nums[i]);
                res[1] = i;
                break;
            }
        }
        
        return res;
    }
}

AC C++:

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> map;
        for(int i = 0; i < nums.size(); i++){
            if(map.find(target - nums[i]) != map.end()){
                return {map[target - nums[i]], i};
            }else{
                map[nums[i]] = i;
            }
        }

        return {-1, -1};
    }
};

跟上Two Sum II - Input array is sorted, Two Sum III - Data structure design, Two Sum IV - Input is a BST, 3Sum, 3Sum Closest, 4Sum.