LeetCode【1】-Two Sum JAVA

题目

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.

也就是一个数组以及一个值，返回数组中两个数相加等于指定值的两个索引

example：

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

方法1

暴力索引，时间复杂度为O(n^2)。

public class TwoSum {/** * 时间复杂度为O(n^2) * @param nums * @param target * @return */public static int []twoSum1(int []nums,int target){    int index1,index2;    int []index = new int[]{0,1};    for(int i = 0;i<nums.length;i++){        for(int j=i+1;j<nums.length;j++){            if(target ==(nums[i]+nums[j])){                index[0]=i;                index[1]=j;                return index;            }        }    }    return index;}

方法2

使用HashMap来做，首先检验target-nums[i]能否加入到HashMap中，若能，则说明前面的数据没有与第i个字符的组合，当添加不成功，则说明存在符合的组合，记录索引，时间复杂度为O（n）。时间缩短很多

public static int []twoSum2(int []nums,int target){    int []index = new int[]{0,1};    HashMap<Integer, Integer> hm = new HashMap<>();    for(int i=0;i<nums.length;i++){        if(hm.containsKey(target-nums[i])){            index[1] = i;            index[0] = hm.get(target-nums[i]);            return index;        }else {            hm.put(nums[i], i);        }    }    return index;}