Two Sum

1、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, 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].

2、求解一（O(n2)）

 * @param nums
     * @param target
     * @return
     *//*
    public int[] twoSum(int[] nums, int target) {
        int len =  nums.length;
        int[] indices = new int[2];
        for(int i = 0; i < len; i++){
            for(int j = i + 1;j < len; j++){
                if(nums[i] + nums[j] == target){
                    indices[0] = i;
                    indices[1] = j;

                }
            }
        }
        return indices;

3、求解二（O(n)）

 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 + 1);
        }
        return result;
    }

此方法的亮点在于只遍历一次，每次遍历利用hashmap将每个数对应的下标和值都存储起来