Leetcode:Two Sum

**1. Two Sum(Easy)**

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.

Example:

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

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

return [0, 1].

public static int[] TwoSum1(int[] nums, int target)
        {
            for (int i = 0; i < nums.Length; i++)
            {
                for (int j = i + 1; j < nums.Length; j++)
                {
                    if (nums[i] + nums[j] == target)
                    {
                        return new int[] { i, j };
                    }
                }
            }
            return null;
        }
        public static int[] TwoSum2(int[] nums, int target)
        {
            List<int> list = new List<int>();
            for (int i = 0; i < nums.Length; i++)
            {
                list.Add(nums[i]);
            }
            for (int j = 0; j < nums.Length; j++)
            {
                if (list.Contains(target - nums[j]) && list.IndexOf(target - nums[j]) !=  j)
                    return new int[2] { j, list.IndexOf(target - nums[j]) };
            }
            return null;
        }
        public static int[] TwoSum3(int[] nums, int target)
        {
            List<int> list = new List<int>();
            for (int j = 0; j < nums.Length; j++)
            {
                list.Add(nums[j]);
                int completation = target - nums[j];
                if (list.Contains(completation) && list.IndexOf(completation) != j)
                    return new int[2] { j, list.IndexOf(completation) };
            }
            return null;
        }
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值