Problem###
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
Analysis###
关键是寻找两个合适的数
最先以为序列式有序的，结果不是；
然后直接穷举所用组合，结果会超时；
最后使用HashMap来实现，key = 数的值，value = 数的位置：首先将所有的数与其位置存储于Map中，然后对于每个数去检测能和它组合成target的数的位置，如果没有，就下一个。
Tips###
1：numbers 是无序的！
2：穷举会超时！:
3：如果Hash搜索的时候是按numbers的顺序从前往后找的话，其本身就是index1小，index2大。
Implementation###
package twosum; import java.util.HashMap; public class Solution { public int[] twoSum(int[] numbers, int target) { HashMap<Integer,Integer> map = new HashMap<Integer,Integer>(); for(int i = 0 ; i < numbers.length ; i++){ map.put(numbers[i], i); } for(int i = 0 ; i < numbers.length ; i++){ int left = target - numbers[i]; if(map.get(left) != null && map.get(left) != i){ int[] indexs = {(i+1),(map.get(left)+1)}; return indexs; } } return null; } }
