Two Sum

Two Sum

问题：

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.

思路：

　　hashtable

我的代码：

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
         HashMap<Integer,Integer> save = new HashMap<Integer,Integer>();
         HashMap<Integer,Integer> duplicate = new HashMap<Integer,Integer>();
         for(int i = 0; i < numbers.length; i++)
         {
             if(save.containsKey(numbers[i]))
             {
                 duplicate.put(numbers[i],i+1);
             }
             else
             {
                 save.put(numbers[i],i+1);
             }
         }
         int[] rst = new int[2];
         for(int i = 0; i < numbers.length; i++)
         {
             int val = numbers[i];
             int remain = target - val;
             if(val == remain)
             {
                 if(duplicate.containsKey(val))
                 {
                     rst[0] = i+1;
                     rst[1] = duplicate.get(val);
                     return rst;
                 }
                 continue;
             }
             if(save.containsKey(remain))
             {
                 rst[0] = i+1;
                 rst[1] = save.get(remain);
                 return rst;
             }
         }
         return rst;
    }
}

他人代码：

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;
}

学习之处：

• 我的代码里面考虑了那么多的Corner Case又是用两个HashTable进行存储，又是判断是否是一个指啊。
• 别人代码用的是回头望月，便往后走，边看看前面是否符合条件，一遍扫描就OK，一个Hashtable就OK，也没有那么多的Corner Case
• 别人的代码简洁，大气，上档次---简洁之美