leetcode 167. Two Sum II - Input array is sorted

「爱情、让人受尽委屈。」 2022-06-08 07:26 160阅读 0赞

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

这道题直接遍历即可。

代码如下：

/*
     * 由于是排好序的，所以遍历查找即可
     * */
    public class Solution 
    {
        public int[] twoSum(int[] numbers, int target) 
        {
            int []res={
       0,0};
            for(int i=0;i<numbers.length;i++)
            {
                res[0]=i+1;
                int one=target-numbers[i];
                boolean find=false;
                for(int j=i+1;j<numbers.length;j++)
                {
                    if(numbers[j]==one)
                    {
                        find=true;
                        res[1]=j+1;
                        break;
                    }else if(numbers[j]>one)
                    {
                        find=false;
                        break;
                    }
                }
                if(find)
                    break;
            }
            return res;
        }
    }

下面是C++的做法，就是做一次遍历

代码如下：

#include <iostream>
    #include <algorithm>
    #include <climits>
    #include <vector>
    #include <string>
    #include <unordered_map>
    
    using namespace std;
    
    
    class Solution 
    {
    public:
        vector<int> twoSum(vector<int>& numbers, int target)
        {
            vector<int> res(2, 0);
            for (int i = 0; i < numbers.size(); i++)
            {
                res[0] = i + 1;
                int one = target - numbers[i];
                bool find = false;
                for (int j = i + 1; j < numbers.size(); j++)
                {
                    if (numbers[j] == one)
                    {
                        find = true;
                        res[1] = j + 1;
                        break;
                    } else if (numbers[j] > one)
                        break;
                }
                if (find)
                    break;
            }
            return res;
        }
    };