LeetCode 1. 两数之和

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].

Example 2:

Input: nums = [3,2,4], target = 6

Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6

Output: [0,1]

Constraints:

• 2 <= nums.length <= 10^4
• -10^9 <= nums[i] <= 10^9
• -10^9 <= target <= 10^9
   

Only one valid answer exists.

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

解法思路：

1、暴力求解

2、利用map集合的特性

class Solution {
    public int[] twoSum(int[] nums, int target) {
        // 暴力求解，O(n^2)
        // int[] res = 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) {
        //             res[0] = i;
        //             res[1] = j;
        //             return res;
        //         }
        //     }
        // }
        // return new int[0];

        // O(n)
        // Map map = new HashMap<>();
        // int[] res = new int[2];
        // for (int i = 0; i < nums.length; i++) {
        //     int anotherNum = target - nums[i];
        //     if (map.containsKey(anotherNum)) {
        //         res[0] = map.get(anotherNum);
        //         res[1] = i;
        //         return res;
        //     }
        //     map.put(nums[i], i);
        // }
        // return new int[0];

        // 优化下
        Map map = new HashMap<>();
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            int anotherNum = target - nums[i];
            if (map.containsKey(anotherNum)) {
                return new int[]{map.get(anotherNum), i};
            } else {
                map.put(nums[i], i);
            }
        }
        return null;
    }
}