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.-----------------------------------------------------------------------------------------------------------------
1）
由于数组是已经排序好的，所以用二分查找，时间复杂度为O(nlogn)。就是先遍历数组，然后查找这个数的右边的是否有一个数，这个数与它相加得到目标数。
C++代码：
class Solution { public: vector<int> twoSum(vector<int>& numbers, int target) { for(int i = 0;i < numbers.size(); i++){ int t = target - numbers[i],left = i + 1,right = numbers.size() - 1; while(left <= right){ int mid = left + (right - left)/2; if(numbers[mid] == t) return {i+1,mid+1}; else if(numbers[mid] < t) left = mid + 1; else right = mid - 1; } } return {}; } };
2）
不过二分查找的时间复杂度比较大，可以用双指针，时间复杂度为线性。空间复杂度为O(1)。
C++代码：
class Solution { public: vector<int> twoSum(vector<int>& numbers, int target) { int l = 0,r = numbers.size() - 1; while(l < r){ if(numbers[l] + numbers[r] == target) return {l+1,r+1}; else if(numbers[l] + numbers[r] > target) r--; else l++; } return {}; } };转载于:https://www.cnblogs.com/Weixu-Liu/p/10785692.html
