在一組陣列中找出兩個數，其加總恰等於給定值。 每個數不能被重複使用，且必剛好只有一個解。
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].
1.暴力破解: 兩個迴圈遍例相加，解法要算N(N-1)/2次，O(N²)的時間複雜度。
public class Solution {
public int[] TwoSum(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 nums;
}
}
2.使用Dictionary紀錄已比較過的item:
找尋Dictionary中是否有存在target減去nums中所遍歷的item差值，如果有即回傳，沒有則將item的值與index存入Dictionary。
優點為每個item數值只會被比較過一次，最壞的狀況為遍歷整個array後，就可以得到答案，由於Dictionary的ContainsKey與Add皆為O(1)，所以最糟的情況下遍歷整個Dictionary需要O(N)。
public class Solution {
public int[] TwoSum(int[] nums, int target) {
Dictionary<int, int> dic = new Dictionary<int, int>();
for (int i = 0; i < nums.Length; i++)
{
var diff = target - nums[i];
if (dic.ContainsKey(diff))
{
return new int[] { dic[diff], i };
} else
{
if (!dic.ContainsKey(nums[i])) {
dic.Add(nums[i], i);
}
}
}
return nums;
}
}
Ref:
https://medium.com/@desolution/%E5%BE%9Eleetcode%E5%AD%B8%E6%BC%94%E7%AE%97%E6%B3%95-1-400da76b51b4
https://leetcode.com/articles/two-sum/
