LetCode 算法题 TwoSum

package letcode.com;

import java.util.HashMap;
import java.util.Map;
/**
 *  1. twoSum 

    Given an array of integers, return indices of the two numbers such that they add up to a specific target.

    You may assume that each input would have exactly one solution, and you may not use the same element twice.
    给定一个整数数组，返回这两个数字的索引，使它们合计成一个特定的目标。

    您可能会认为每个输入都只有一个解决方案，并且您可能不会使用相同的元素两次。

    Example:

    Given nums = [2, 7, 11, 15], target = 9,

    Because nums[0] + nums[1] = 2 + 7 = 9,
    return [0, 1].

 * @author gfw2306
 *
 */
public class TwoSum {


    /**
     * 此方法时间复杂度为O(n) ，只有一层遍历。
     * @param nums
     * @param target
     * @return
     */
    public int[] twoSum(int[] nums, int target) {
        long begin = System.currentTimeMillis();
        int[] result = new int[2];
        Map<Integer,Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < nums.length; i++) {
            if(!map.containsKey(nums[i])){
                map.put(nums[i], i);
            }
        }
        A:for (int i = 0; i < nums.length; i++) {
            int b = target - nums[i];

            if(map.containsKey(b) && i != map.get(b)){
                result = new int[]{i,map.get(b)};
                break A;
            }
        }
        long end = System.currentTimeMillis();
        System.out.println("复杂的为O(n)的方法耗时： "+(end - begin)+"毫秒");
        return result;
    }

    /**
     * 此方法使用循环套循环方式，典型的二维矩阵，时间复杂度为O(n^2)
     * @param nums
     * @param target
     * @return
     */
    public int[] twoSumMethodOne(int[] nums, int target) { 
        long begin = System.currentTimeMillis();
        int[] answer = new int[2]; 
        A:for (int i = 0; i < nums.length; ++i){
            answer[0] = i; 
            int b = target - nums[i]; 
            for (int j = i + 1; j < nums.length; ++j){ 
                if (nums[j] == b){ 
                    answer[1] = j; 
                    break A; 
                } 
            } 
        }
        long end = System.currentTimeMillis();
        System.out.println("复杂的为O(n^2)的方法耗时： "+(end - begin)+"毫秒");
        return answer;
    }






    public static void main(String[] args) {
        TwoSum twoSum = new TwoSum();
        int[] nums = new int[]{1,12,3,4,5,6,8,9,12,2,3,11,12,34,7,8};
        int[] result = twoSum.twoSum(nums, 4);
        System.out.println(result[0]+"--"+result[1]);

        int[] result2 = twoSum.twoSumMethodOne(nums, 4);
        System.out.println(result2[0]+"--"+result2[1]);
    }



}