two-sum

packagecom.ytx.hash;

importjava.util.Arrays;

importjava.util.HashMap;

/**

 *  题目：two-sum

 * 

 *  描述：    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

                    

                    

 *@authoryuantian xin

 * 以下不同时间复杂度的做法

 *

 *     1.很直观的做法，O(N2)做法：

 *

 *  publicint[] twoSum(int[] numbers, int target) {

             int len = numbers.length;

             if(len < 0) return null;

             

             int[]  res = newint[2];

             

             for(int i = 0; i <len; i++) {

                    for(int j = i + 1; j <len; j++) {

                           if(numbers[i] + numbers[j] == target) {

                                 res[0] = i + 1;

                                 res[1] = j + 1;

                                 break;

                           }

                    }

             }

       

             return res;

    }

   

 * 

 *  2.  O(n)  哈希表。将每个数字放在map中，历遍数组，如果出现和数组中的某一个值相加为target的时候，break。

 *  这个方法同样适用于多组解的情况。

 * 

 * int n = numbers.length;

       int[] result = new int[2];

        //map里面放 键为target-numbers[i] 值为下标i

        //每次放入的时候看是否包含 当前值

        //有的话说明当前值和已包含的值下标的那个元素为需要的结果

        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();

        for(int i=0;i<n;i++){

            if(map.containsKey(numbers[i])){

                result[0] = map.get(numbers[i])+1;

                result[1] = i+1;

                break;

            }

            else{

                map.put(target- numbers[i], i);

            }

        }

        return result;

       

       

    3.最优O(nlogn)。排序，然后两个指针一前一后。因为题中说明了只有一对答案，因此不需要考虑重复的情况。

 *

 */

publicclass Two_sum {

       

       

       publicint[] twoSum(int[]numbers,int target) {

             intlen = numbers.length;

       int[]result = newint[2];

        Arrays.sort(numbers);//数组升序排序

       int left = 0;

       int right = len - 1;

       while(left< right) {

             //如果相等直接得到结果，跳出循环

             if(numbers[left] + numbers[right] ==target) {

                    result[0] =left + 1;

                    result[1] =right + 1;

                    break;

                    

             }else if(numbers[left] + numbers[right] <target) {

                    //因为数组升序，如果left和right的和小于target，左指针右移

                    left++;

                    

             }else {

                    ///因为数组升序，如果left和right的和大于target，右指针左移

                    right--;

             }

             

        }

       

       return result;

    }

       publicstatic void main(String[]args) {

             intnumbers[] = {2, 7, 11, 15};

             inttarget = 9;

             intres[] =new int[2];

             res= new Two_sum().twoSum(numbers,target);

             

             System.out.println(res[0]);

             System.out.println(res[1]);

       }

}