1.Two sum (Easy)

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].

solution

题意是给定一个int数组nums和一个整数target，在nums里面找到两数之和为target的数，并返回两数的数组索引。假定nums数组里面只有唯一的一组结果，且同一个数字只能使用一次。

1.最容易想到的暴力算法，也就是我起初想到的！！！

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] indices = new int[2];
        for (int i = 0; i < nums.length - 1; i++)
        {
            for (int j = i + 1; j < nums.length; j++)
            {
                if (nums[i] + nums[j] == target)
                {
                    indices[0] = i;
                    indices[1] = j;
                    return indices;
                }
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

2.速度更快的解法，HashMap

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

reference
https://leetcode.com/articles/two-sum/

总结

第一种暴力算法就是先取一个数，然后用这个数依次与后面的每一个数相加求和，若和与target相等，则将这两个数的数组索引加入到一个新的int数组中，然后返回这个int数组。time complexity:O(n^2),space complexity:O(1)

原文地址：https://www.cnblogs.com/victorxiao/p/11135576.html