Two Sum

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.

题意：
给出一个整数数组和一个数target，求出两个下标i,j,使得a[i]+a[j] = target,返回下标

Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

思路:
1.最简单的两重循环，选定数组中的一个数a[i],再遍历数组，看是否有target-a[i],返回下标即可,时间复杂度为O(n2).
2.利用Java中的HashMap,循环遍历数组中的每个元素，利用HaspMap进行查找target-a[i],HaspMap的查找的时间复杂度为O(1),因此总的时间复杂度为O(n).

代码:

public class TwoSum {    /*    O(n2)     */    public int[] twoSum(int []nums,int target){        int solution[] = new int[2];        for(int i = 0; i < nums.length; i++){            int temp = nums[i];            for(int j = 0; j < nums.length; j++){                if(j == i) continue;                if(nums[j] == target-nums[i]){                    if(i > j){                        solution[0] = j;                        solution[1] = i;                    }else{                        solution[0] = i;                        solution[1] = j;                    }                    i = nums.length+1;                    break;                }            }        }        return solution;    }    /*    O(n)     */    public int[] twoSum2(int []nums,int target){        int solution[] = new int[2];        Map<Integer,Integer> map = new HashMap<Integer, Integer>();        for(int i = 0; i < nums.length; i++){            if(map.containsKey(target-nums[i])){                solution[1] = i;                solution[0] = map.get(target-nums[i]);                return solution;            }            map.put(nums[i],i);        }        return solution;    }    public static void main(String[] args) {        int nums[] = {2,7,11,15};        int solution[] = new TwoSum().twoSum2(nums,9);        System.out.println(solution[0] + " " + solution[1]);    }}