思路

方法一：二分

遍歷每個數字num，然後再在後面的數字中使用二分查詢target-num。

複雜度分析

時間複雜度：O(nlogn)

空間複雜度：O(1)

 1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int>& nums, int target) {
 4         int left, mid, right;
 5         for(int i = 0; i < nums.size()-1; ++i) {
 6             left = i+1
 
;
 7             right = nums.size()-1;
 8             while(left <= right) {
 9                 mid = (left + right) / 2;
10                 if(nums[mid] == target-nums[i])
11                     return vector<int>({nums[i], nums[mid]});
12                 if(nums[mid] < target-nums[i]) {
 
13                     left = mid + 1;
14                 } else {
15                     right = mid - 1;
16                 }
17             }
18         }
19 
20         return vector<int>();
21     }
22 };

方法二：雙指標(對撞指標)

 1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int
 
>& nums, int target) {
 4         int i = 0, j = nums.size()-1;
 5         while(i < j) {
 6             int s = nums[i] + nums[j];
 7             if(s < target) ++i;
 8             else if(s > target) --j;
 9             else return vector<int>({nums[i], nums[j]});
10         }
11 
12         return vector<int>();
13     }
14 };