LeetCode算法入门- Two Sum-day1

Two Sum

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]

方法一：
时间复杂度0(n^2):
利用两个for循环来遍历数组，暴力破解法

class Solution {
    public int[] twoSum(int[] nums, int target) {
        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 null;
    }
}

方法二：时间复杂度为0(n),
通过构造一个HashMap，充分利用HashMap.containsKey()和HashMap.get(key)和HashMap.put(key,value)方法来完成题目。从下标为0开始遍历，如果HashMap中没有符合条件的值时，就将（数组当前的值，数组当前的下标）（key，value）放到HashMap当中。

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] res = new int[2];
        Map map = new HashMap<>();
        for(int i = 0; i < nums.length; i++){
            int result = target - nums[i];
            if(map.containsKey(result)){
                res[0] = i;
                res[1] = map.get(result);
                return res;
            }
            else{
                map.put(nums[i],i);
            }
        }
 return null;
    }
}