LeetCode 167. Two Sum II - Input array is sorted(有序数组夹逼法)

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本文介绍了一种在已排序的整数数组中寻找两个数使其和等于特定目标值的方法。提供两种算法实现：一种使用双重循环，另一种采用两端逼近的方式，有效地解决了问题并降低了时间复杂度。

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Total Accepted: 74198
Total Submissions: 157252
Difficulty: Easy
Contributor: LeetCode

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：

1.使用两层for循环，因为数组为有序数组，所以当两数之和大于target时，要break，从而减少比较次数，降低时间复杂度。

2.使用夹逼法，设置两个索引r,l

Code1：

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        for(int i=0;i<numbers.size();i++){
            for(int j=i+1;j<numbers.size();j++){
                int sum =numbers[i]+numbers[j];
                if(sum>target) break;
                if(sum == target){
                    vector<int> index{i+1,j+1};
                    return index;
                } 
            }
        }
    }
};

Code2：

vector<int> twoSum(vector<int>& numbers, int target) {
        int l = 0;
        int r = numbers.size() -1;
        while(l < r){
            if(numbers[l] + numbers[r] == target){
                vector<int> res{l+1,r+1};
                return res;
            }
            else if(numbers[l] + numbers[r] > target){
                r--;
            }
            else{
                l++;
            }
        }
    }