【LeetCode】 两数之和 II – 输入有序数组

• 2019 年 12 月 3 日
• 筆記

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本文链接：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.

有个大数会卡一下，先看看我写的（辣鸡死了）

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，两头堵~~~~就不是n2超时了

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;      }  };