【LeetCode】 兩數之和 II – 輸入有序數組
- 2019 年 12 月 3 日
- 筆記
版權聲明：本文為部落客原創文章，遵循 CC 4.0 BY 版權協議，轉載請附上原文出處鏈接和本聲明。
本文鏈接：https://blog.csdn.net/shiliang97/article/details/103188270
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.
Note:
- 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.
Example:
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.
有個大數會卡一下，先看看我寫的（辣雞死了）
while(numbers[i]==numbers[i+1]&&numbers[i]*2!=target)i++;
class Solution { public: vector<int> twoSum(vector<int>& numbers, int target) { vector<int> v; for(int i=0;i<numbers.size();i++){ while(numbers[i]==numbers[i+1]&&numbers[i]*2!=target)i++; int t=target-numbers[i]; for(int l=i+1;l<numbers.size();l++){ if(numbers[l]==t){ v.push_back(i+1); v.push_back(l+1); return v; } } }return v; } };
人家其他人，兩頭堵~~~~就不是n2超時了
class Solution { public: vector<int> twoSum(vector<int>& numbers, int target) { vector<int> index; int l=0; int r=numbers.size()-1; while(l<r) { if(numbers[l]+numbers[r]==target) { index.push_back(l+1); index.push_back(r+1); return index; } else if(numbers[l]+numbers[r]<target) ++l; else --r; } return index; } };
