LeetCode From Easy To Hard No1[Easy]: Two Sum

Description: 

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].

 

一开始想到的是用两重for遍历数组每个元素

import java.util.Arrays;

public class Solution {
    public int[] twoSum(int[] nums, int target) {
  
        int[] result = 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) {
                    result[0] = i;
                    result[1] = j;
                }
            }
        }
        return result;
    }
    public static void main(String[] args) {

        int[] nums = {2, 7, 11, 15}; int target = 9;

        long startTime=System.nanoTime(); //获取开始时间

        int[] result = new Solution().twoSum(nums, target);

        long endTime=System.nanoTime(); //获取结束时间

        System.out.println("程序运行时间： "+(endTime-startTime)+"ns");

        System.out.println(Arrays.toString(result));

    }
}

上边的solution套用了两重循环，时间复杂度为O(m*n), 后来参考别人的代码发现更好的solution - 利用map只需要套一层循环可以做到边存边查，时间复杂度为O(n)：

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        int[] result = new int[2];
        for (int i = 0; i < nums.length; i ++) {

            if (map.get(target - nums[i]) != null) {
                result[0] = map.get(target - nums[i]);
                result[1] = i;
            } else {
                map.put(nums[i], i);
            }

        }
        return result;
    }
    public static void main(String[] args) {

        int[] nums = {2, 7, 11, 15}; int target = 9;

        long startTime=System.nanoTime(); //获取开始时间

        int[] result = new Solution().twoSum(nums, target);

        long endTime=System.nanoTime(); //获取结束时间

        System.out.println("程序运行时间： "+(endTime-startTime)+"ns");

        System.out.println(Arrays.toString(result));

    }
}