心得

這題我本來是想用兩個for迴圈解決的

public class Solution {
    public int[] TwoSum(int[] numbers, int target) {
        for (int i = 0; i < numbers.Length; i++)
        {
            for (int j = numbers.Length - 1; j > i; j--)
            {
                if (numbers[i] + numbers[j] == target)
                {
                    return new int[] { i + 1, j + 1 };
                }
            }
        }

        throw new Exception();
    }
}

但發現會Time Limit Exceeded於是乎看了一下Solutions他們是用一個while迴圈來解答，題目要求index1<index2且不能使用重複的元素，所以這題如果用while的話宣告兩個變數分別是left與right就可以只用一個迴圈就解決。

問題

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution and you may not use the same element twice.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

答案

public class Solution {
    public int[] TwoSum(int[] numbers, int target) {
        int left = 0;
        int right = numbers.Length - 1;
        while (numbers[left] + numbers[right] != target)
        {
            if (numbers[left] + numbers[right] > target)
            {
                right--;
            }
            else
            {
                left++;
            }
        }

        return new int[] { left + 1, right + 1 };
    }
}