【LeetCode】- Two Sum（两数相加）

[ 问题: ]

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

题意: 给定一个整数, 从里面找出两个数, 其和等于一个指定的整数. 程序返回这两个数在数组中的位置( 数组下标从1开始 ) , 且位置小的在前面.

[ 常规: ]

下面的解法时间复杂度为O(n * m), 也是很多人第一反应就能得出的解法, 但是非常的遗憾, 这个解法无法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帮我们记住了值和索引, 对每一个数边查边存, 时间复杂度为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]);
		}
	}
}