【LeetCode】C# 1、Two Sum
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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].
这是LeetCode第一题,我从这一题开始入坑。
刚看到这一题,刚接触算法的我第一想法是直接遍历如下,时间复杂度O(n2),很直接,很暴力,很不给过。。。当LeetCode给出一个很大的case的时候超时了。
public class Solution { public int[] TwoSum(int[] nums, int target) { int[] ret = new int[2]; for (int left = 0; left < nums.Count() - 1; left++) { for (int right = left + 1; right < nums.Count(); right++) { if (nums[left] + nums[right] == target) { ret[0] = left; ret[1] = right; } } } return ret; }}
之后我学聪明了一点,换成左右两个指针,分别向中间靠拢,复杂度O(n),然而,我忽略了一点,nums[]是无序的。这个思路就不行了,只能重写。
public class Solution { public int[] TwoSum(int[] nums, int target) { int[] ret = new int[2]{0,0}; int left=0,right=nums.Count()-1; while(left<right){ if(nums[left]+nums[right]==target){ ret[0]=left; ret[1]=right; } else if(nums[left]+nums[right]>target) right--; else left++; } return ret; }}
最后,用一个dictionary来保存各个数据的位置。遍历一遍,知道出答案为止。case中最后一个坑也出在这里。。这个超长的case中,居然有两个863,结果报错,添加了两个重复的keys。还好这次不用重写,添加了个if解决问题。
最终方案:
public class Solution { public int[] TwoSum(int[] nums, int target) { int[] ret = new int[2]; Dictionary<int, int> dict = new Dictionary<int, int>(); for (int i = 0; i < nums.Length; i++) { if (dict.ContainsKey(target - nums[i])) { ret[1] = i; ret[0] = dict[target - nums[i]]; return ret; } if(!dict.ContainsKey(nums[i])) dict.Add(nums[i], i); } return ret; }}
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