题目描述 LeetCode 167

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.

题目解读

• 从一个数组中找到两个数，使这两个数相加等于给定的target，输出数组中这两个数的下标(下标从1开始)

解题思路

• 题目中规定，第一个数的下标小于第二个；所以依次遍历数组，如取出第 i 个数字，则将 i 后面的数字作为一个新的数组放入二分查找中进行搜索，搜索目标为 target - number[i], 如果找到则返回下标，循环终止；找不到则返回-1，开始下一次循环

Code

# include<stdio.h>

int BinarySearch(int rr[], int left, int right, int x){
    int index = (right + left) / 2;
    int mid = rr[index];

    if(left <= right){
        if(x < mid){
            return BinarySearch(rr, left, index - 1, x);
        }
        else if(x > mid){
            return BinarySearch(rr, index + 1, right, x);
        }
        else{
            return index;
        }
    }
    else{
        return -1;
    }
}

int* twoSum(int* numbers, int numbersSize, int target, int* returnSize) {
    int i;
    int index1, index2;
    static int temp[2];

    for (i = 0; i < numbersSize; i ++){
        index1 = i;
        index2 = BinarySearch(numbers, i + 1, numbersSize - 1, target - numbers[i]);
        if (index2 > 0){
            break;
        }
    }
    temp[(*returnSize) ++ ] = index1 + 1;
    temp[(*returnSize) ++ ] = index2 + 1;
    return temp;
}

int main(){ 
    int a[10] = {0, 0, 3, 4};
    int target = 0;
    int returnSize = 0;
    int i;
    int* temp;

    temp = twoSum(a, 4, target, &returnSize);

    for(i = 0; i < returnSize; i ++){
        printf("%d  ", temp[i]);
    }
}

• 思路二 二分查找 C++ 实现

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        vector<int> result;
        
        for(int i=0; i < numbers.size() - 1; i++){
            int tt = BinarySearch(numbers, i+1, numbers.size()-1, target-numbers[i]);
            if(tt > 0){
                result.push_back(i+1);
                result.push_back(tt+1);
                break;
            }
        }
        return result;
    }
    
    int BinarySearch(vector<int>& nums, int left, int right, int target){
        if(left <= right){
            int indexOfmed = (left + right) / 2;
            int med = nums[indexOfmed];
            
            if(target < med){
                return BinarySearch(nums, left, indexOfmed-1, target);
            }
            else if(target > med){
                return BinarySearch(nums, indexOfmed+1, right, target);
            }
            else{
                return indexOfmed;
            }
        }
        else{
            return -1;
        }
    }
};

思考