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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:
代码语言：javascript复制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 ;
代码语言：javascript复制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
代码语言：javascript复制
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;
}
};
