2 sum

2 sum

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

Note

You may assume that each input would have exactly one solution

Example

numbers=[2, 7, 11, 15], target=9

return [1, 2]

Challenge

1. O(1) Space, O(nlogn) Time

2. O(n) Space, O(n) Time

Solution:

O(1) Space, O(n^2) Time

    public int[] twoSum(int[] numbers, int target) {        int[] result = new int[2];        if (numbers == null || numbers.length == 0) {            return result;        }        for (int i = 0; i < numbers.length; i++) {            for (int j = i + 1; j < numbers.length; j++) {                if (numbers[i] + numbers[j] == target) {                    result[0] = i + 1;                    result[1] = j + 1;                    return result;                }            }        }        return result;    }

O(n) Space, O(n) Time

    public int[] twoSum(int[] numbers, int target) {        int[] result = new int[2];        if (numbers == null || numbers.length == 0) {            return result;        }        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();        for (int i = 0; i < numbers.length; i++) {            map.put(numbers[i], i + 1);        }        for (int i = 0; i < numbers.length; i++) {            if (map.containsKey(target - numbers[i])) {                result[0] = i + 1;                result[1] = map.get(target - numbers[i]);                break;            }        }        return result;    }

思路：

1. 嵌套遍历，由于原数组非有序，不能提前终止循环。

2. 占用额外空间换取查找速度，利用哈希表（数值，下标加1），快速定位下标。