1. Two Sum

1. Two Sum (easy)

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.

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

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

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(1)解题思路 o(n2)

使用遍历的方式，取数组中的每一个元素，遍历其后的元素与其相加是否等于目标值。

public class Solution {    public int[] twoSum(int[] nums, int target) {        int index[] = new int[2];        int i=0, j=0;        for (i = 0; i < nums.length; i++) {            for (j = i+1; j < nums.length; j++) {                if(target == nums[i]+nums[j]){                    index[0] = i;                    index[1] = j;                    return index;                }            }        }        return null;    }}

(2)解题思路o(n) 参考的

使用hashmap将值与索引一一映射，同时考虑到了当数组中有重复元素时的情况
eg: nums[] = {0,3,5,0} target = 0
此时键值不是一一对应的，如果key不唯一,而hash(key)是单值函数,就会使得后面添加进来的value覆盖前面的key对应的value，这道题在循环时i是从0开始的，所以这里不会影响结果
代码精炼，时间复杂度低，值得学习

public class Solution {    public int[] twoSum(int[] nums, int target) {        HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();        for(int i = 0; i<nums.length;i++){            m.put(nums[i], i);        }        for(int i=0; i<nums.length; i++){            if(m.containsKey(target-nums[i]) && m.get(target-nums[i])!=i)                return new int[]{i, m.get(target-nums[i])};        }        return new int[]{0};    }}

(3)参照上面，可以使用ArrayList实现o(n)

在使用这种方式时，由于数组中有重复元素，所以返回结果时应该要对数组进行重新的排序

 ArrayList <Integer> list = new ArrayList<>();        for (int i = 0; i < nums.length; i++) {            list.add(nums[i]);        }        for (int i = 0; i < list.size(); i++) {            if(list.contains(target-list.get(i)) && list.indexOf(target-list.get(i))!=i){                int index[]=new int[]{i,list.indexOf(target-list.get(i))};                Arrays.sort(index);                return index;            }        }        return  new int[]{0};