LeetCode 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].

此題通過甚爲簡單，本人Accepted代碼如下：

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] A = new int[2];
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] + nums[j] == target) {
                    A[0] = i;
                    A[1] = j;
                    break;
                }
            }
            if (A[0] + A[1] == target) {
                break;
            }
        }
        return A;
    }
}

此方案內存比99.68%的提交者少，但rt僅比18.77%快，故看了看其他人的解法。

一種爲O(n)的解法如下：

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

@jiaming2