leetcode題目之兩數求和

題目很簡單，返回數組中等於目標的兩數的下標。用了兩種方法去求解。第一種兩層循環，時間複雜度O（n^2） 第二種用了hashmap。時間複雜度O（n）.
question:
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].

解題 法1：

    public static int[] twoSum(int[] nums, int target) {
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[j] == target - nums[i]) {
                    return new int[] { i, j };
                }
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }

法2：

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

測試：

public static void main(String[] args)
    {
        int[] nums={1,0,4,8};
        twoSum(nums,9);
        int[] num1=twoSum(nums,9);
        System.out.println(Arrays.toString(num1));

    }

輸出：[0,3]