1、two sum

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Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

给一个整形数组，从里面找出两个相加等于target的数字，并输出其index

由于一开始认为给定的数组是未排序的，并且用最笨的方法来实现：

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

时间复杂度为O(n^2)，RumTime为372 ms

后来根据其给出的提示，优化的思路为：因为给定的整数数组中，我们要找的两个数字分别为x和target-x。因此，创建一个hashmap，利用hashmap的key的唯一性，将每一位数字所对应的target-x值作为key，index为value，直到找到了另一个数字在hashmap中存在着value。

public class Solution {    public int[] twoSum(int[] nums, int target) {        int[] index = new int[2];        HashMap<Integer,Integer> hashMap = new HashMap<>();        for(int i=0;;i++){            if(hashMap.containsKey(nums[i])){                index[0] = hashMap.get(nums[i]) +1;                index[1] = i + 1;                if(index[0]>index[1]){                    int temp = index[0];                    index[0] = index[1];                    index[1] = temp;                }                break;            }else                hashMap.put(target-nums[i],i);        }        return index;    }}

上面代码偷了一下懒，因为假如给定的数组中必定存在两个值，则一定能输出结果，所以for循环里就缺省了判断范围，否则可能分分钟报错

至此，时间复杂度为O(n)，RunTime为 6 ms

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