O(n2) runtime, O(1) space – Brute force:

The brute force approach is simple. Loop through each element x and find if there is another value that equals to target – x. As finding another value requires looping through the rest of array, its runtime complexity is O(n2).

O(n) runtime, O(n) space – Hash table:
We could reduce the runtime complexity of looking up a value to O(1) using a hash map

that maps a value to its index.

Solution in Java

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

Solution in Python

class Solution(object):
    def twoSum(self, nums, target):
        # two point
        nums_index = [(v, index) for index, v in enumerate(nums)]
        nums_index.sort()
        begin, end = 0, len(nums) - 1
        while begin < end:
            curr = nums_index[begin][0] + nums_index[end][0]
            if curr == target:
                return [nums_index[begin][1], nums_index[end][1]]
            elif curr < target:
                begin += 1
            else:
                end -= 1
 if __name__ == '__main__':
    # begin
    s = Solution()
    print s.twoSum([3, 2, 4], 6)