Leedcode 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, 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].

算法：

   In the first iteration, we add each element's value and its index to the table. Then, in the second iteration we check if each element's complement ($target-nums[i]$) exists in the table. Beware that the complement must not be $nums[i]$ itself!

    While we iterate and inserting elements into the table, we also look back to check if current element's complement already exists in the table. If it exists, we have found a solution and return immediately.

  设存在着一对解m, n，设m在序列中排在n的前面。从头到尾遍历该序列，用数据结构S存储已经遍历过的数据。遍历到m时候，寻找target-m是否在S中，这时候还是不在。但是遍历到n的时候就发现target-n，也就是m，在S中，这时候我们就得到了解。该方法的时间复杂度为O(n)，但是可能需要额外O(n)的存储空间。而数据结构要很方便于查询，所以可使用字典（C++中用map，Python中用dict，存储数据以及下标），查询复杂度为O(1)。

  注意点在于map的索引是给定nums值，存放的数据是下标。

  比如给3，2，4，Target为6。设map为空，当第一迭代，i=1取到3时，6-3=3不在map中，将3存为索引，1存为3对应的值。当第二迭代，i=2取到2时，6-2=4不在map中，将2存为索引，2存为2对应的值。当第三次迭代，i=3取到4时，6-4=2在map中，返回i和map中索引为6-4=2的数值。

python:

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        tmp_num = {}  
        for i in range(len(nums)):  
            if target - nums[i] in tmp_nums:  
                return (tmp_num[starget-nums[i]], i)  
            else:  
                tmp_num[nums[i]] = i  
        return (-1, -1)  

C++：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        map<int, int> searched;  
        vector<int> res;  
        for (int i = 0; i < nums.size(); ++i){  
            if (searched.count(target - nums[i])){  
                res.push_back(searched[target - nums[i]]);  
                res.push_back(i);  
                return res;  
            } else {  
                searched[nums[i]] = i;  
            }  
        }  
        res.push_back(-1);  
        res.push_back(-1);  
        return res;  
    }
};

java:

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No two sum solution");
}