LeetCode01:Two Sum

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

这里我就不做翻译了，我这里给出几种做法，有些参考了一些资料

First Method:不排序处理

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        Set processedNumbers = new HashSet();
        boolean foundSum = false;
        int[] result = new int[2];
        for(int i=0;i<numbers.length;i++){
            int reqNumber = target-numbers[i];
            System.out.println("current number =" +numbers[i]+",requiredNumbers="+reqNumber);
            if(processedNumbers.contains(reqNumber)){
                result[0] = numbers[i];
                result[1] = reqNumber;
                foundSum = true;
                break;
            }else{
                processedNumbers.add(numbers[i]);
            }
        }
        if(!foundSum){
            throw new RuntimeException("Sum is not found");
        }
        return result;
    }
}

Second Method:排序后再处理

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
       Arrays.sort(numbers);
       int length = numbers.length;
       int left = 0;
       int right = length-1;
       int[] result = new int[2];
       int sum = 0;
       while(left<right){
           sum = numbers[left]+numbers[right];
           if(target==sum){
               result[0]=numbers[left];
               result[1]=numbers[right];
           }else if(target<sum){
               if(left<right)
               right--;
           }else if(target>sum){
               if(left<right)
               left++;
           }
       }
       return result;
       }
}

Third Method:类似快排原理

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
       //Arrays.sort(numbers);
       int length = numbers.length;
       int left = 0;
       int right = length-1;
       int[] result = new int[2];
       int sum = 0;
       while(left<right){
           sum = numbers[left]+numbers[right];
           if(target==sum){
               result[0]=numbers[left];
               result[1]=numbers[right];
           }else if(target<sum){
               if(left<right)
               right--;
           }else if(target>sum){
               if(left<right)
               left++;
           }
       }
       return result;
       }
}

Fourth Method:感觉比较巧妙的java函数处理

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;
    }
}

个人觉得能通过oj，但是却没有，别人能通过，感觉可能是rp有问题，这里就简单分享一下