[ 問題: ]

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.

You may assume that each input would have exactly one solution.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

題意: 給定一個整數, 從裡面找出兩個數, 其和等於一個指定的整數. 程式返回這兩個數在陣列中的位置( 陣列下標從1開始 ) , 且位置小的在前面. 

[ 常規: ]

下面的解法時間複雜度為O(n * m), 也是很多人第一反應就能得出的解法, 但是非常的遺憾, 這個解法無法AC.

Status: Time Limit Exceeded

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

	public static void main(String[] args) {
		int[] numbers = { 2, 7, 11, 15 };
		int target = 9;
		int[] result = new Solution().twoSum(numbers, target);
		for (int i = 0; i < result.length; i++) {
			System.out.println(result[i]);
		}
	}
}
 

[ 正解: ]

下面的巧妙的利用map幫我們記住了值和索引, 對每一個數邊查邊存, 時間複雜度為O(n)

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

	public static void main(String[] args) {
		int[] numbers = { 2, 7, 11, 15 };
		int target = 17;
		int[] result = new Solution().twoSum(numbers, target);
		for (int i = 0; i < result.length; i++) {
			System.out.println(result[i]);
		}
	}
}