題目來源
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
思路就是利用哈希表军援，記錄遍歷到的秦陋，查詢target減去當前的map是否存在蜜葱，直接看代碼，O(n)的復雜度谬运，感覺沒有用到sorted這個一特性，應該還有更好的方法。
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int n = numbers.size();
vector<int> res;
unordered_map<int, int> map;
for (int i=0; i<n; i++) {
if (map[target - numbers[i]]) {
res.push_back(map[target - numbers[i]]);
res.push_back(i+1);
return res;
}
map[numbers[i]] = i + 1;
}
return res;
}
};
然后發(fā)現(xiàn)自己傻逼了赌蔑，根本沒有想的那么復雜丸相，只要兩個指針搔确，從兩端向中間掃描，然后就可以了灭忠。
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int n = numbers.size();
int l = 0, r = n - 1;
while (l < r) {
if (numbers[l] + numbers[r] < target)
l++;
else if (numbers[l] + numbers[r] > target)
r--;
else
return vector<int>({l+1, r+1});
}
return vector<int>();
}
};
