leetcode: Two Sum II - Input Array is Sorted

• 1. 題目
• 2. 解答
• 3. 總結

1. 題目

Given a 1-indexed array of integers numbers that is already sorted in 
non-decreasing order, find two numbers such that they add up to a specific target 
number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <=
index1 < index2 <= numbers.length.
Return the indices of the two numbers, index1 and index2, added by one as an 
integer array [index1, index2] of length 2.
The tests are generated such that there is exactly one solution. You may not use 
the same element twice.
Your solution must use only constant extra space.
           

Example 1:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].
           

Example 2:

Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].
           

Example 3:

Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].
           

Constraints:

• 2 <= numbers.length <= 3 * 104
• -1000 <= numbers[i] <= 1000
• numbers is sorted in non-decreasing order.
• -1000 <= target <= 1000
• The tests are generated such that there is exactly one solution.

2. 解答

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int left = 0, right = numbers.size() - 1;
        while(left < right)
        {
            if(numbers[left] + numbers[right] == target)
                return {left + 1, right + 1};
            else if(numbers[left] + numbers[right] > target)
                --right;
            else
                ++left;
        }
        return {-1, -1};
        
    }
};
           

   雙指針解法，可以了解為一個“元解法"。它基于排序的數組進行。利用大、小關系做一些确定性結論，進而縮小範圍。

  注意“Given a 1-indexed array”，是以傳回結果要在0-indexed的基礎上+1。

3. 總結

  部分解法或思想需要以常識或者直覺的方式記憶在腦中，我這邊稱之為"元解法"。就好比數學中的乘法口訣表的存在。

  對應題幹上異常的說明，要劃出來。在方案寫完之後，對應要點，是否都已在代碼的考慮之内。