Leetcode 1: Two Sum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Solution:

1. Brute Force

   a. 这是我第一反应的解法，常规的循环遍历相加

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] ans = new int[2];
        for (int i = 0; i <= nums.length - 2; ++i) {
            for (int j = i + 1; j <= nums.length - 1; ++j) {
                if (nums[i] + nums[j] == target) {
                    ans[0] = i;
                    ans[1] = j;
                    return ans;
                }
            }
        }
        throw new IllegalArgumentException("no match found");
    }
}

  b. 这种暴力搜索方法是做减法，更容易引申下面HashMap的解法

class Solution {
    public int[] twoSum(int[] nums, int target) {
        for (int i = 0; i < nums.length; ++i) {
            for (int j = i + 1; j < nums.length; ++j) {
                int complement = target - nums[i];
                if (nums[j] == complement) {
                    return new int[] {i, j};
                }
            }
        }
        throw new IllegalArgumentException("no match found");
    }
}

2.  HashMap

HashMap存储nums数组的值和角标的键值对，找到则返回，找不到则存入HashMap

此种解法时间复杂度为O(n)

注意，HashMap containsKey方法的时间复杂度为O(1)，containsValue方法的时间复杂度为O(n)

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