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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
import java.util.HashMap; import java.util.Vector; public class Solution { public int[] twoSum(int[] numbers, int target) { int result[]= new int[2]; HashMap<Integer,Vector<Integer>> aaa = new HashMap<Integer,Vector<Integer>>(); for(int i=0;i<numbers.length;i++) { if( aaa.containsKey(numbers[i]) ) { Vector<Integer> index = aaa.get(numbers[i]); index.add(i+1); } else { Vector<Integer> index = new Vector<Integer>(); index.add(i+1); aaa.put(numbers[i], index); } } for(int i=0;i<numbers.length;i++) { int subTarget = target - numbers[i]; if(aaa.containsKey(subTarget)) { Vector<Integer> index = aaa.get(subTarget); for(int j = 0;j<index.size();j++) { if(index.get(j)!=i+1) { result[0] = i+1; result[1] = index.get(j); return result; } } } } return result; } }
上面是个很笨的解法，下面是Discuss里面一个很酷的解法：
import java.util.Hashtable; public class Solution { public int[] twoSum(int[] numbers, int target) { int[] result = new int[2]; Hashtable<Integer, Integer> table = new Hashtable<Integer, Integer>(); for(int i = 0; i < numbers.length; i++){ if(table.containsKey(numbers[i])){ result[0] = table.get(numbers[i]) + 1; result[1] = i + 1; } table.put(target - numbers[i], i); } return result; } }
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原文地址：http://www.cnblogs.com/wzm-xu/p/4342966.html
