leetcode Two Sum

leetcode

1、Two Sum

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.

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

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

翻译过来的意思就是说要在数组中找到能够满足两个数的和刚好等于target的值得位置，注：数组下标是从0开始的。

方法一：直接暴力搜：

    public class Solution {
        public int[] twoSum(int[] nums, int target) {
            int len=nums.length;
            int a[] = new int[2];
            int tag=1;
            for(int i=0;ifor(int j=i+1;jif((nums[i]+nums[j])==target)
                    {
                        a[0]=i;
                        a[1]=j;
                        tag=0;
                        break;
                    }
                }
                if(tag==0)
                break;
            }
            return a;
        }
    }

方法二：上面的方法可能会报超时的错误，所以我们可以换个姿势重新过，用HashMap,使得时间复杂度将为：O(n)（因为一般情况下，HashMap的时间复杂度为O(1)）

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        int len=nums.length;
        Map map = new HashMap();
        int a[] = new int[2];
        int tag=1;
        for(int i=0;ifor(int i=0;iint j = target - nums[i];
            if(map.get(j)!=null && !map.get(j).equals(Integer.valueOf(i)))
            {
                a[0] = i;
                a[1] = (int)map.get(j);
                break;
            }
        }
        return a;
    }
}