题目大意：给你一个排过序的数组，让你输出两个数的index，他们的和等于target。
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
Solution:
C++ / two pointers
Time complexity: O(n)
Space complexity: O(1)
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// Author: Huahua // Running time: 6 ms class Solution { public: vector<int> twoSum(vector<int>& numbers, int target) { int i = 0; int j = numbers.size() - 1; while (i < j) { const int sum = numbers[i] + numbers[j]; if (sum == target) break; else if (sum < target) ++i; else --j; } return {i + 1, j + 1}; } }; |
C++ / Binary search
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// Author: Huahua // Running time: 6 ms class Solution { public: vector<int> twoSum(vector<int>& numbers, int target) { const int n = numbers.size(); for (int i = 0; i < n; ++i) { int l = i + 1; int r = n; int t = target - numbers[i]; while (l < r) { int m = l + (r - l) / 2; if (numbers[m] == t) return {i + 1, m + 1}; else if (numbers[m] < t) l = m + 1; else r = m; } } return {}; } }; |
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